
Oass__JxLLlQ. 

Book i_L4<a- 

CopigM 



COPYRIGHT DEPOSIT. 



AN ELEMENTARY TREATISE 

ON 

THE MECHANICS OF MACHINERY 



'?&&& 



AN ELEMENTARY TREATISE 



ON THE 



MECHANICS OF MACHINERY 

WITH SPECIAL REFERENCE TO THE 

MECHANICS OF THE STEAM-ENGINE 



BY 
y 
Y 

JOSEPH N. LeCONTE 

INSTRUCTOR IN MECHANICAL ENGINEERING, UNIVERSITY OF CALIFORNIA 
ASSOCIATE MEMBER OF THE AMERICAN INSTITUTE 
OF ELECTRICAL ENGINEERS, ETC. 



Nefo fgork 
THE MACMILLAN COMPANY 

LONDON: MACMILLAN & CO., Ltd. 
1902 

All rights reserved 



TTHC 



THE LIBRARY OF 

CONGRESS, 
TVo Cowts Received 

DFC, 6 190? 

OLASwO^tXXc No. 

cory b. 



Copyright, 1902, 
BY THE MACMILLAN COMPANY. 



Set up and electrotyped November, 1902. 




Norroooti $ress 

J. S. Cushing & Co. - Berwick & Smith 
Norwood Mass. U.S.A. 



P 



PREFACE 

o 



^ The following book is the outcome of a course of lectures on 
^Kinematics and the Mechanics of the Steam-engine, which has 
k^been issued in the form of notes to students in the Department 
of Mechanical Engineering of the University of California for 
several years past. The first two parts embody the more im- 
portant principles of what is generally called the Kinematics of 
Machinery, though in many instances dynamic problems which 
present themselves are dealt with ; the real purpose of the book 
being the application of the principles of mechanics to certain 
problems connected with machinery. In the third part there is 
discussed the Mechanics of the Steam-engine, that machine 
being perhaps the most important from a designer's point of 
view. Here the subject is treated under two distinct heads, 
Kinematics and Dynamics. 

No special originality is claimed for the major part of the 
material, and free use has been made of whatever literature 
could be found relating to the special subjects under considera- 
tion. To the more important of these, references are given by 
foot-notes. 

JOSEPH N. Le CONTE. 

Berkeley, California. 
November, 1902. 



CONTENTS 

PART I 

PAGE 

INTRODUCTORY . 3 

PART II 

MACHINERY OF TRANSMISSION 

CHAPTER I 

Transmission of Pure Rotation through a Rigid Inter- 
mediate Member 

1. Axes of Rotation in One and the Same Straight Line ... 21 

A. Direct Connection, Rigid Couplings 21 

2. Axes of Rotation Parallel 23 

A. Parallel Bars or Cranks 23 

B. Oldham's Coupling 24 

C. Certain Forms of the Universal Joint ..... 25 

3. Axes of Rotation Intersecting 25 

A. The Universal Joint 25 

4. Axes of Rotation Crossing 31 

CHAPTER II 

Transmission of Pure Rotation by Means of Friction 
Gearing 

1. Axes of Rotation in One and the Same Straight Line ... 32 

A. Direct Connection, Friction Couplings 32 

2. Axes of Rotation Parallel 32 

A. Ordinary Friction Wheels 32 

3. Axes of Rotation Intersecting, and 35 

4. Axes of Rotation Crossing 35 

vii 



viii CONTENTS 



CHAPTER III 

Transmission of Pure Rotation by Means of Belts and 
Ropes 

PAGE 

1. Axes of Rotation Parallel, Planes of the Pulleys the Same . . 36 

A. Ordinary Belting 36 

(a) Stresses in the Belt, and Power transmitted ... 37 

(b) Stepped Cones 43 

B. Wire Ropes 54 

(a) Horizontal Transmission . . . . . 55 

(&) Inclined Transmission 58 

2. Axes of Rotation Parallel, Planes of the Pulleys Different ... 62 

3. Axes of Rotation Intersecting 64 

4. Axes of Rotation Crossing 65 



CHAPTER IV 

Transmission of Pure Rotation by Means of Toothed 
Gearing 

1. General Considerations 66 

2. Axes of Rotation Parallel, Spur Gears 66 

A. Laws of Action . 66 

B. Velocity Ratio Constant, Circular Wheels .... 72 

(a) Cycloidal System 73 

(&) Involute System 85 

(c) Pin-tooth System 94 

(d) Special Forms of the Above, Twisted Gears . . .100 
0) Strength of Spur Teeth . . . . . .104 

C. Velocity Ratio Variable, Non-circular Wheels . . . .105 
(a) Elliptic Gears 106 

3. Axes of Rotation Intersecting, Bevel Wheels 114 

4. Axes of Rotation Crossing, Skew Wheels 117 

A. Spiral Gears • • lI 7 

B. Worm Gears I 3S 

C. Hyperboloidal Gears . , , , . . . • J 37 



CONTENTS ix 

CHAPTER V 

Transmission of Rectilinear Translation 

PAGE 

i. Prismatic Guides 142 

2. Parallel, or Straight Line Motions 142 

A. Classification 142 

B. Cycloidal Straight Line Motions 142 

C. Conchoidal Straight Line Motions 147 

D. Lemniscate Straight Line Motions 149 

E. Inversors 150 



CHAPTER VI 

Transmission of Motion by Direct Contact when Direc- 
tional Relations are not Constant 

Cams 155 

A. Disk Cams 155 

B. Cylindrical Cams 158 



PART III 

MECHANICS OF THE STEAM-ENGINE 

CHAPTER I 

Kinematics 

1. General Description of the Chain ■ . 163 

2. The Piston-crank Chain 163 

A. Relation between Position of Crank and Positions of Other 

Points in the Chain 164 

B. Relation between Position of Crank and Velocities of Other 

Points in the Chain 167 

C. Relation between Position of Crank and Accelerations of 

Other Points in the Chain 173 



X CONTENTS 

PAGE 

3. Valve Gearing 180 

A. Relation between Position of the Crank and Position of the 

Valve 180 

O) The Plain Slide Valve 180 

(&) The Gridiron Valve 208 

(V) The Meyer Valve 211 

(d) The Thompson Valve 216 

B. Relation between Position of the Crank and Velocities and 

Accelerations of the Valve 218 

4. Relation between Positions of the Piston and Positions of the Valve 219 

CHAPTER II 

Dynamics 

1. Forces due to Steam Pressure in the Cylinder 222 

2. Inertia Effects of the Reciprocating Parts ...... 229 

A. Analytical and Graphical Calculation 229 

B. Applications 239 

(a) Fly-wheel Analysis 239 

(b) Counterbalancing . . 248 

3. Governors 256 

A. Fly-ball Governors 256 

B. Shaft Governors 258 



PART I 

INTRODUCTORY 



INTRODUCTORY 

Kinematics. — Kinematics is that branch of Mechanics which 
deals with motion without reference to the cause producing it. It 
is important in dealing with machinery, where motion is entirely 
independent of the direction of the acting forces. In the follow- 
ing chapters we always assume rigid bodies, that is, bodies such 
as have their particles at an invariable distance from one another. 
Though there are no absolutely rigid bodies, all substances being 
more or less elastic, still, for the materials of machine construc- 
tion, the effect of elasticity in affecting constrained motion is 
insignificant. 

When a body has any motion whatever, its position at any given 
time is completely determined when the position of three of its 
points are known, provided these three points do not lie in one 
straight line. Hence no motion is possible if three such points 
are fixed. If one point is fixed, a knowledge of the position of 
two others, not in line with the first, is sufficient to determine the 
position of the body. Likewise if two are fixed, a knowledge of 
the position of a third only is required. 

Kinds of Motion. — If a body has none of its points fixed, we 
study its motion generally by investigating the paths traced by 
three of its points. Such cases are not frequent in machinery, 
with the exception of helical screw motion, which can be reduced 
to a conbination of two simpler forms. If a body has one point P 
fixed, all its other points move on the surfaces of a system of con- 
centric spheres with centre P, hence this motion is called Spheric 
Motion. In its general form it is not common in machine move- 
ments, but in the special case, where the fixed point is removed to 
infinity, it becomes the most important of all forms. In this case 
the system of concentric spheres becomes a system of parallel 

3 



4 INTRODUCTORY 

planes within any finite space. This form is called Uniplanar 
Motion, and all planes of the body at right angles to the line 
drawn toward the fixed point continue to move in their own 
planes. This fixing of one point at infinity vastly simplifies the 
motion, and nearly all machine motions are of this type. In treat- 
ing of plane motion we have to consider the motion of but one of 
these planes, which may be called the Reference Plane, and the 
motion of all other points of the body can be studied by their pro- 
jection on this plane. Hence the whole is reduced to a problem 
in Plane Geometry. Since one point is already fixed, we have 
only to consider the motion of two points, that is, of a line of the 
body. In uniplanar motion, then, we can treat all bodies as lines. 

Kinds of Uniplanar Motion. — If we still further constrain the 
body by fixing a point in the reference plane or at any finite dis- 
tance from it, we have the particular case of Pure Rotation. 
Here all points in the body describe a system of concentric circles 
about the line connecting the two fixed points, which line is called 
the Axis of Rotation. If the second fixed point is also removed 
to infinity in any direction not at right angles to the reference 
plane, we have the particular case of Rectilinear Translation. 
The concentric circles within any finite space become a system of 
straight lines parallel to the line of intersection of the two refer- 
ence planes. 

Combination of the Two Elementary Forms of Uniplanar Motion. 
— The combination of two or more rectilinear translations is an 
elementary principle of Mechanics, as is also the combination of 
two rotations. A displacement caused by a translation and a rota- 
tion taken in either order can be replaced by a single rotation. 
The construction for this is shown in Fig. i, where changes of 
position only are considered. Let the rotation be such that if act- 
ing alone the body would move from AB to A'B' about C as a 
centre. The translation in the same plane is such as would shift 
the body from A'B' to A"B". The result of the two is to shift 
the body from AB to A"B". Connect AA" and BB" by straight 
lines, and at the middle points of these last erect perpendiculars 
intersecting at /, which then must be the point sought. Now if 



INTRODUCTORY 



5 




the body moves according to a given law, we may take a series 
of determinate positions, and find the point / for each successive 
pair of these. Finally, when these successive positions become 
indefinitely close together, the 
motion becomes continuous, 
and hence the centre / must 
move continuously also. Since, 
then, the position of the centre 
/ at any time has but an instan- 
taneous value, it is called the 
Instantaneous Centre. 

The Instantaneous Centre. — 
When the positions AB and 
A"B" are indefinitely close to- 
gether, the arc A A" and the 
chord A A" coincide, and the 
line IE coincides with IA, and 
is perpendicular to v 1} the direc- 
tion in which A is moving. The 

same is true with reference to IB, IF, and z/ 2 . Hence if we know 
at any instant the direction of motion of two points of a body 
(moving with uniplanar motion), the instantaneous or virtual 
centre will lie at the intersection of the perpendiculars drawn to 
them. 

Centrodes of Motion. — The instantaneous centre will itself, 
therefore, trace out a curve on the reference plane. The curve 
thus traced out with reference to axes fixed in space is known as 
the Space Centrode, or Fixed Centrode. If we refer the motion 
of the instantaneous centre to axes fixed in the moving body, we 
shall obtain another curve called the Body Centrode, or Moving 
Centrode: The body centrode can be constructed for any given 
position of the body by transferring every other position of the 
body to the given position, and carrying with each its instanta- 
neous centre. Another curve is thus formed which must have a 
point in common with the first ; namely, the instantaneous centre 
of the reference position. A simple pair of centrodes which cor- 



6 INTRODUCTORY 

respond to a line moving with its extremities on a pair of rectan- 
gular coordinates is shown in Fig. 2. Let AB be one position of 

the body. A moves ver- 
tically downwards, and B 
horizontally to the right. 
The instantaneous centre, 
therefore, lies at the inter- 
section of the perpendic- 
ulars to the directions of 
motion of these, or at /. 
Other positions of the 
body give p lt p 2 , etc. 
These points lie on a cir- 
cle with centre O, and 
radius 01 = AB, which 
is the space centrode. 
Choosing any reference 
position of the body such 
as AB, we construct the body centrode by replacing on AB 
every other position of the body with its instantaneous centre 
" attached." For example, the position A 2 B 2 with centre at p 2 
gives, when replaced on AB, the point q 2 , on the" body centrode. 
The body centrode in this case is seen to be a circle of half the 
diameter of the space centrode, since it is the locus of the right 
angle of a right triangle of invariable hypothenuse. When a 
circle rolls within another of double its diameter, every point on 
the circumference of the small circle describes a diameter of the 
large one, which is a straight line hypocycloid. The pair of cen- 
trodes above is a pair of such circles. But A and B (points on 
the circumference of one circle) move on diameters of the other 
circle by the conditions of the problem ; hence in this particular 
case the centrodes roll without slipping. Moreover, this is true 
for all cases.* Another more complex example of a pair of cen- 
trodes is shown in Fig. 3, and exhibits the case of the connecting 




Fig. 2 



For proof of the general theorem of rolling centrodes, see Appendix I. 



INTRODUCTORY 7 

rod of an ordinary steam engine. Here both are infinite curves 
with straight asymptotes. Any point such as F (Fig. 2) is mov- 
ing at right angles to IF; hence if about / as a centre, with a 
radius equal to IF, we describe a circular arc, it will touch the 
path of F. But the same is true of a circle drawn about p Y with 



.- ~JY 




Fig. 3 



a radius q x F. Thus if we draw a series of arcs about p lf p. 2 , p g , etc., 
with radii equal to q x F, q. 2 F, q 3 F, etc., these will form the envelope 
of the path of F.* 

Applications of the Instantaneous Centre in Machines. — In de- 
scribing a machine we will use the definition given by Professor 



* For important and interesting properties of rolling curves, see Appendix II. 



8 INTRODUCTORY 

Kennedy* as being the best, and it is as follows : "A machine is 
a combination of resistant bodies, whose relative motions are com- 
pletely constrained, and which serve by these relative motions to 
transform the energies at our command into any special form of 
work." Considering the different parts of the definition, he points 
out that : — 

i. A machine is a combination of bodies ; that is, no one body 
can constitute a machine. 

2. They must be resistant bodies, or bodies which have the 
property of resisting force in the direction in which it is applied. 
In the great majority of cases we deal with practically rigid 
bodies, or bodies which have their particles at an invariable 
distance from one another. 

3. The motion must be constrained, or all points when referred 
to a given set of reference axes must follow paths determined by 
the construction of the machine itself, and not by the direction 
and magnitude of the acting forces. 

4. Motion is an essential condition, as otherwise energy will not 
be transformed. 

We are now to examine the methods used to constrain relative 
motion. We must have in the first place at least a pair of bodies 
or elements. The motion, if completely constrained, will depend 
on the form of the portions in contact. If these portions are so 
constructed that the contact between the bodies at all times during 
motion is a surface contact, the pair is known as a Lower Pair. 
But if it is such that the bodies touch upon a point or along a line, 
they are called Higher Pairs. 

Elementary Machine Motions. — The two commonest and most 
important forms of motion to be found in machinery are the two 
elementary forms of uniplanar motion, viz., the rotation and the 
rectilinear translation. The first of these can be realized when the 
parts in contact are identical portions of any surfaces of revolu- 
tion, and the second when they are identical prismatic surfaces. 
It will be noticed that the nature of both of these surfaces is such 

* "Mechanics of Machinery," Alex. B. Kennedy. Macmillan & Co. 



INTRODUCTORY 9 

as to form a lower pair of elements, and while from a geometric 
standpoint the higher and lower pair are equally good for the con- 
strainment of motion, practically the lower is the preferable as 
being less affected by the wearing away of the parts. The vast 
majority of machine motions are constrained in these ways, as by 
the journal and bearing, and the prismatic guide. The only other 
surface which will work as a lower pair is the cylindric screw sur- 
face, and this does not constrain a uniplanar motion. 

Relative Motion. — In order that all points of a machine should 
be completely constrained, no part should move without all others 
undergoing a corre- 
sponding change in 
position. Take, for 
example, the com- 
bination of pairs or 
links shown in Fig. 4. 
Consider the body 
a fixed in space, so 
that the motions of 

the other three can be examined relatively to it. The links b and 
c are then completely constrained by being hinged at O and P to 
a, since all points of these links must follow definite, i.e. circular, 
paths. Concerning the link d we know that it has a point R common 
j to b, which is therefore constrained relatively to a, and a point Q 




Fig. 




common to c similarly constrained. Therefore the body d is com- 
pletely constrained, as the paths of two points determine the 
uniplanar motion of a body. It will be noticed that the motions 
of b and c relatively to a are pure rotations, but that the 
motion of d is not. If any other link besides a were fixed, we 



10 



INTRODUCTORY 



would arrive at a similar result, hence the combination shown in 
Fig. 4 is a true machine. Fig. 5 is another very common example 
of a machine, being in fact the elementary combination used in 




Fig. 6 

the steam engine. Fig. 6 is, however, not a machine, as neither 
of the links x or y are constrained relatively to a. If the point A 
is connected to a by means of a sixth link z, x and y are then con- 
strained. 

Determination of the Instantaneous Centre in Machines. — In 
order to determine the instantaneous centre of a link referred to 




Fig. 7 

any other link, we must know the direction of motion of two of its 
points at the position considered. In Fig. 7 we have the same 
combination of links as in Fig. 4. Here, as in all cases, we can 



INTRODUCTORY 



II 



write down at once the permanent centres as O ai , O ac , O bd , and 
O cd , the notation meaning that, for example, O ab is the centre 
about which b is turning relatively to a (or a to b), etc. But d Ts 
not turning about a fixed centre relatively to a, and only an instan- 
taneous centre exists for this motion. In order to determine O ad , 
we know that O bd (a point common to both b and d) is moving in 
a direction v 1 perpendicular to the line O ab O bd , and that the 
instantaneous centre O ad lies somewhere on a line at right angles 
to this direction of motion, or somewhere on the line O ah O bd , pro- 
duced if necessary. Similarly we know it must lie somewhere on 
O ac O cd . Therefore it lies at the intersection of these two as 
shown in the diagram. The same result would evidently be 
obtained if we were to study the motion of a relatively to d. By 
exactly the same process of reasoning we find, when either b or c 
are fixed, that the relative instantaneous centre of these two lies 
at the intersection of O ab O ac and O bd O cd . We have in all six 
centres (permanent or instantaneous), which give the relative 
points of rotation of every link with respect to every other. If 
we select any three links in the above 
mechanism, such as a, b, and d, the three 
relative centres are O ah , O ad , and O bd , 
and it will be noticed that these lie in a 
straight line, as do the relative centres of 
any other three links. 

That this law is universally true may 
be shown as follows : let a, b, and c, be 
any three bodies having relative plane 
motion, whose three centres are at O ab , 
O ac , and O bc (Fig. 8). Consider a fixed. 
Then b is turning about O ab , and c about 
O ac . But according to the definition of 
a centre, the point O bc has no motion 
relatively to either b or c, hence it is a Fig. 8 

point common to both. Considered as 

a point of b, it is moving in a direction Y, at right angles to 
O ab O bc , and when considered a point of c it is moving in a 




12 



INTRODUCTORY 



direction X, at right angles to O ac O bc , both taken relatively to a. 
But, as a point can move in but one direction relatively to a 
given body, X and Y must coincide, or O bc must lie on the line 
joining O ac and O al . This proposition is of great help in finding 
the relative centres of more complicated combinations of links 
than Fig. 7, or in those where the relative motions are not easily 
seen by mere inspection. In the simple combination reproduced 
in Fig. 9 the centre O bc can be found by its aid. Fig. 10 




Fig. 9 



shows the fifteen centres of a more complicated combination 
of six links. The above law can be traced throughout the 
construction. 

Determination of Relative Linear Velocity. — A most useful 
property of the instantaneous centre in machine design is that 
which enables us to find the velocity of any point in the machine 
when that of any other is given. Let a, b, and c (Fig. n) be any 
three bodies, of which a is fixed. The three relative centres will 
evidently lie in a straight line. Let v Y be the known velocity of any 
point P in b. It is required to find the velocity v 2 of point Q in 



INTRODUCTORY 



13 



'Obf 




Fig. 10 



the body c. The velocities of all points of a rotating body being 
proportional to their respective distances from the centre of 
rotation, we have : 



YelO bc = v 1 ^ 



O nh P 



the velocities being referred to a, and 
O bc being a point of b. But also : 



Vel. O lc = v 9 Q^> 
O ac Q 



where O bc is a point of c. As O bc is a 
point common to both b and c, its 




Fig. 11 



14 



INTRODUCTORY 



velocity, when considered a point of either and referred to a i 
must be the same. Hence : 



„ _ 7; O ah O b O ac Q 



O ah P o ac o bc 

A simple graphical solution of the above equation is shown in Fig. 
n. This consists of finding by circular projection points R and S 
on the line O ab O ac which have the same velocities as P and Q, 




Fig. 12 

and solving the equation by similar triangles. Suppose, in the 
mechanism of Fig. 10, we have given the velocity of a point P 
(Fig. 12) in link d, to find velocity of point Q in link e, both 
referred to a. We first pick out from Fig. 10 the three relative 
centres O ad , O ae , and (le , all of which are instantaneous centres. 
Pis turning about O ad relatively to a, hence the velocity of P will 
be v x of magnitude given, and in the direction at right angles to 
PO„ d . We then project Pow a circular arc to R, on the straight 



INTRODUCTORY 



15 



line joining the three centres. From R we lay off the velocity of 
P in any direction, though it is preferable to lay it off at right 
angles to the line of centres, in order to preserve the direction as 
well as the magnitude of the velocity of R also. Connect the 
extremity of Vi with O ad , and draw v through O de parallel to vj to 
intersect this last line. v is the velocity of O de , or the common 
point of the two moving bodies. Q is turning about O ae , hence 




Fig. 13 

project Q on a circular arc to .Son the line of centres. Through 
the extremity of v and O ae draw a line and produce in the direction 
of S. Through 5 draw v£ parallel to v x ' to intersect this last line. 
Then v 2 ' is the velocity of .S in magnitude and direction, and the 
required velocity of Q, viz., v 2 is drawn at right angles to QO ae , 
and equal in magnitude to v 2 '. The above construction applies in 
all cases except where all three centres are at infinity. It is equally 
applicable whether the points under consideration lie in the same 



i6 



INTRODUCTORY 



or in different links, and whether any or all of the links are turning 
about permanent or instantaneous centres. When the points lie 
in the same link, the two centres O ae and O ad coalesce, and O de , 
the point common to both, may be any point of the moving body. 
In this case the line connecting the three virtual centres and upon 
which we project the points /'and Q is not fixed in direction, and 
hence may be taken in any direction at random. It is preferable 
to take it passing through one of the points under consideration, 




C d 



Fig. 14 



as by doing so the construction is simplified. Fig. 13 shows the 
simple construction in this case, both where the projection line is 
drawn at random and where it is drawn through Q. 

In the case where the three relative centres lie at infinity, the 
motion of the two links which carry the points must be motions of 
translation relatively to the fixed link, and relatively to each other. 
But in this case all points in the moving bodies have the same 
velocities, and therefore the points of the links hinged to these have 



INTRODUCTORY 



17 



the same velocities also. So the problem reduces to the compar- 
ison of the velocities of these points of attachment. Take the case 
exhibited in Fig. 14, which is essentially the reducing motion of the 
Tabor Indicator. We have two blocks, a and b, moving parallel 
to one another, with rectilinear translations, but with different veloc- 
ities, which we wish to compare. This is evidently the same thing 
as comparing the velocity of the point O ad of the link d with that 
of O he of e. All the centres, as well as the construction in the par- 
ticular case, are shown. 

Determination of Relative Angular Velocity. — Problems in 
Angular Velocity can be solved in a similar manner by the use of 
the instantaneous centre. Referring to 
Fig. 15, let the body b turn with an an- 
gular velocity a^ about O ah , required <o 2 , 
that of c about O ac . O bc being a point 
common to both b and c, we have already 
seen that its velocity referred to a, is the 
same when considered a point of either. 
Calling its velocity v , we have 



or 



w 2 — w l ~ZZ — TT" 




Fig. 15 



This is easily solved graphically by laying 
off in any direction a^, the given angular 
velocity of b from the centre of c referred 
to a, then drawing a line from its extremity through O hcy and pro- 
ducing if necessary to meet a line from the centre of b parallel to 
coj. This last line will give the magnitude of <o 2 , and evidently, if O bc 
lies between the other two, the angular velocities will be in oppo- 
site directions, and if not they will be in the same direction. The 
above proposition stated in words is as follows : the relative 
instantaneous centre of two moving bodies divides the distance 
between their centres in the inverse ratio of their angular veloci- 
ties referred to a fixed body. Applying this in Fig. 16 to the 
c 



i8 



INTRODUCTORY 



combination of Fig. 10, we have the angular velocity of d about 
its instantaneous centre given, and are required to find that of e. 
The given angular velocity o) x is laid off from O ae) and <o 2 is drawn 




Fig. 16 



parallel to it and from O ad to meet the line from the tip of o^ 
through O de . Generally the only angular velocities which are 
of importance in machine design are those taking place about 
permanent centres. 



PART II 

MACHINERY OF TRANSMISSION 



CHAPTER I 

TRANSMISSION OF PURE ROTATION THROUGH A RIGID 
INTERMEDIATE MEMBER 

i. AXES OF ROTATION IN ONE AND THE SAME 
STRAIGHT LINE 

A. Direct Connection and Rigid Couplings 

In this case mereTy an axle or shaft is used to connect the two 
rotating parts, and the kinematics of the arrangement is of the 
utmost simplicity. Regarding the dimensions of the parts, how- 
ever, a few words may be necessary. 

Journals and Bearings. — A Journal in machinery is a rigid 
body bounded by a surface of revolution, which is usually cylindric 
or conic. In order that this piece may partake of no motion 
other than rotation, a second piece must be associated with it ; 
namely, the Bearing. This must be bounded by the same surface 
of revolution with concavities and convexities reversed. The 
journal and bearing, therefore, form a pair of elements designated 
as a turning pair, and the character of the contact is that of a 
lower pair. Owing to the frequency with which such journals 
occur in machinery, they require careful designing. This should 
first be done according to the principles of the Strength of Mate- 
rials. If we consider the journal as an overhanging beam of 
length / uniformly loaded, the equation of moments will be 

**■ Pl KI 

M= — = — , 
2 e 

where K is a constant of the material, /the moment of inertia of 
a circle about its diameter, and e the distance of the outermost 
fibre from the neutral layer. 



22 MACHINERY OF TRANSMISSION 

o- t "xd* , d 

Since /=— - and e = -, 

64 2 

e 32 

, />/ *d 3 

and — = K 

2 32 

This involves two variables, /and ^/j but as we can generally assume 
their ratio, we may write : . 



c 
d 



whence d= 2. 26a/ 






d being the diameter in inches. (For cast iron, ^=4200; 
wrought iron, ^=8500; steel, A^= 12000.) 

A journal designed in this way for strength may be, and gen- 
erally will be, found much too small to stand the wear of con- 
tinuous use and radiate the heat produced by friction at a 
sufficiently rapid rate. In this case it is usual to allow a certain 
pressure per square inch of projected area, this latter being the 
product of length and diameter. Let this allowable pressure be/. 
The projected area is / x d, and / x d x p = P. Now as /= c x d, 



-t 



The value of p varies greatly, depending on the total load and the 
nature and velocity of the rubbing surfaces. Its limits are gen- 
erally between 100 and 500 pounds per square inch.* 

When considerable space intervenes between journals, the 
connecting part may be called an Axle. If loaded it also must be 
designed according to Strength of Materials. 

Shafting. — When of great length and used for transmitting 
power, the shaft and its journals are to be designed to resist tor- 
sion also. This is given by the well-known formula : 

* For accurate design of journals, see some work on Machine Design. 



AXES OF ROTATION PARALLEL 23 

, slC x B.P. 
. * = \— -N-* 

where C is a constant of the material, N the number of revolu- 
tions per minute, and H.P. the horse-power transmitted. 
The following are rough values of C : 

Wrought-iron shafting . . . . C ' = 1 00 

Steel shafting C = 75 

Also for shafting up to 4" in diameter, 
Z = 4.8 a/7 2 , 

where L is the distance between hangers in feet, and d the 
diameter of the shaft in inches. 

Pivots are end journals which sustain pressure in the direction 
of their axes. Their diameters need be calculated to withstand 
wear and heating only. They are usually flat on the thrust sur- 
face, though for light, high-speed machinery they are sometimes 
conical. For heavy thrusts, collars are turned on the shaft. 

A line of shafting is often of very great length, and is made of 
many pieces coupled together. This gives rise to many forms of 
rigid couplings, but this is purely a problem of Machine Design. 

2. AXES OF ROTATION PARALLEL 

A. Parallel Bars, or Cranks 

In the combination of links shown in Fig. 4 a transmission of 
rotation is effected between parallel shafts at O and P. In order 
that both shafts should revolve through a complete circle, it is 
necessary that the two links b and c must be of the same length, 
as must a and d also. It will be noticed in this case, however, 
that the motion is not constrained at the instant the four per- 
manent centres come in line. To overcome this defect a second 
pair of cranks must be placed upon the same shaft, making some 
angle other than zero or 180 with the first pair. This second 
pair is usually at 90 with the first pair, and the combination is well 
illustrated in the side rods of locomotives. 



24 



MACHINERY OF TRANSMISSION 



B. Oldham's Coupling 

In this form, two disks A and B (Fig. 17) are keyed to the ends 
of the shafts. These disks are connected by means of an inter- 



E* 



y 



Fig. 17 

mediate disk C, upon each of whose faces a prismatic groove is 
cut, the two crossing one another at an angle (3. These grooves 

fit two prismatic ridges, one on 
each of the disks A and B, and 
thus driving is effected. If the 
shaft A turns at any angular 
velocity, B will follow at the 
same velocity. In Fig. 18, A 
and B are the centres of the 
shafts, and the lines DD and 
EE intersecting at C are the 
*">» ridges of the disks. As the 
disks turn, these lines will al- 
ways pass through A and B, and 
on account of the intermediate 
disk C they will always intersect 
at the same angle /3. Hence the locus of C is a circle on AB as 
a chord. Let .£>'£>' and EE' be any other positions of the ridges. 
Then angle Z>BZ>' = angle EAE' or A will turn through the same 




Fig. 18 



AXES OF ROTATION INTERSECTING 



25 



angle as B. It will be noticed that the motion of the intermediate 
disk C can be reproduced by the rolling upon circle ABC of 
another circle of double the diameter. (Compare Fig. 2.) Driv- 
ing is best effected when the angles at C are right angles. Then 
AB becomes a diameter. 

C. Certain Forms of the Universal Joint 

This will be described under intersecting shafts in the next 
section. 

3. AXES OF ROTATION INTERSECTING 

A. The Universal Joint 

Hook's Universal Joint is used between intersecting shafts when 
it is necessary to vary the angle between the shafts. The single 
joint can be used only between intersecting shafts, but the double 
joint can be used between parallel shafts also, and this latter com- 
bination forms the most flexible of all couplings. 




Its construction is as follows : the shafts are provided with 
forked ends AB and CD (Fig. 19), whose four bearings at A, B, 
C, and D, fit the four journals of an intermediate body. This lat- 
ter is generally in the form of an equal armed cross, a circular disk, 
or a sphere, where the common axes of the opposite pairs AB and 
CD intersect in a point S at right angles. The forks are so pro- 



26 



MACHINERY OF TRANSMISSION 




portioned that S lies at the intersection of the shaft axes also. 
When the shaft F turns, the centre line of its bearings A and B 
moves in a plane at right angles to the axis of F, and will trace out 
a great circle AGBH (Fig. 20) on the surface of a sphere whose 

centre is S. At the same time 
the driven shaft E must turn, 
and the axis of the bearings 
of its fork must remain in a 
plane at right angles to the 
axis of F, describing a great 
circle ADBC on the sphere. 
When the forks of F are at A 
and B, those of E are at D 
and C. W T hen the forks of F 
are at G and N, those of E 
are at A and B ; or when the 
driving fork has turned through 
a quadrant from either of these 
positions, the driven fork turns 
through a quadrant also. Consider the forks of F to have turned 
through an angle a, so that A arrives at O. Then E will turn 
through an angle (3, and D will move to P, where : 
cos OSF= (cos ASO) (cos ASP) + (sin ASO) (sin ASP) (cos OAF), 

cos 90° = cos« cos (90° + /?) -f sin a sin (90°+/?) cos 8, 
where 8 is the angle between the planes of the great circles A GBH 
and AF>BC, which is also the constant inclination between the in- 
tersecting shafts. Hence : 

o = —cos a sin /3+ sin a cos ft cos 8, 

« cos a sin B tan B 

cos 8 = £ = ^. 

sin a cos (3 tan a 

The angles a and ft are therefore not in general equal, nor is their 
ratio constant during any time, but the ratio of their tangents is a 
constant and is equal to the cosine of the angle between the shafts. 
Since cos 8 is always less than unity and is positive in any practical 
construction, tan (3 is always less than tan a. Thus in the first and 



Fig. 20 



AXES OF ROTATION INTERSECTING 2 J 

third quadrants fi < a, while in the second and fourth, /? > a. If 
F is driving at a uniform rate, E will follow at a greater and less 
speed alternately. The relation between the angular velocities for 
any given value of a is determined as follows. Let o) X be the angu- 
lar velocity of F, and <o 2 that of E. Then : 

da , dB 

= wt, and ' 



to!, anu -j- = <o 2 , 
at at 



dft _ sec 2 a cos 8 afo 



I + 


tan 2 « cos 2 


8 <# 






0) l cos 


8 




cos 2 


«(i + tan 2 


a cos ; 


3)' 




(o 1 cos 8 







•*=■* 



i — sin 2 « sin 2 8 

If the driver is turning with a constant angular velocity we can 
determine those values of a at which the angular velocity of the 

follower is a maximum or minimum by putting — — equal to zero. 

dt 



da) 2 _ 2 (Oi cos 8 sin 2 8 (sin a cos a) _ 
dt (i — sin 2 oc sin 2 8) 



= <* 



sin a cos a = o. 

q 

This condition is fulfilled when a is either o, -, ir, or — -. In 

2 2 

order to discriminate between the maxima and minima, substitute 
these angles in — j, and the algebraic sign of the result shows 

that: 

a) 2 ] min . at a = o and a = ir, 

w 2jmax at « = — and a = ^ — 

2 2 

Substituting these values in the original expression for the angu- 
lar velocity <d 2 we find : 

w 2_Jmax. = ~» 

COSO 

<°2]min. = w l C OS 8. 



28 



MACHINERY OF TRANSMISSION 



Thus as the driving fork is turned at a constant angular velocity 

<a h that of the follower will vary between ^ and ^ cos 8. 

cos 8 

If written in terms of the angular velocity ratio, it is evident that 





COi 


at a = o and a = ir, 




(0 2 _ 


max. 




ft). 


7T . ?7T 

at « = - and « = — 




0) 2 


min. 2 2 


Likewise : 






Cl> 2 


I 




nmx . cos 8' 




«1 


= cos 8. 






00 2 


min. 



When 8 is 90 , the angular velocity of the follower will vary be- 
tween zero and infinity, or the joint will fail to work. If 8 is zero, 
the angular velocity ratio is constantly equal to unity. For prac- 
tical purposes 8 should not be larger than 30 . 

If o) x is constant, the angular acceleration of the follower can be 
found for any given value of a as follows : 



o) 2 = 



w l 



cos 8 



1 — sin 2 a sin 2 8 

And since o^ is constant 

do3 2 _2w! cos 8 sin 2 8 sin a cos a da 
dt (1 — sin 2 « sin 2 8) 2 dt 

_ 2 a*! 2 cos 8 sin 2 8 ( sin a cos a) 
(1 — sin 2 cc sin 2 8) 2 

which becomes zero at a = o, -, 71-, and ^. Since the quantity 

2 2 

2 Wi 2 cos 8 sin 8 is a constant, we can find the value of a corre- 
sponding to the maximum or minimum value of — = by simply con- 

dt 
sidering the variation of — sin a cos a — p ut ^jg e q Ua i to/. 



(1 — sin 2 «sin 2 8) : 



AXES OF ROTATION INTERSECTING 



29 



Then: 

(cos 2 <*— sin 2 «)(i — sin 2 a sin 2 8) 2 +4 since cos a sin 2 8 (sin a cos a) 

dp (1— sin 2 6csin 2 8) 2 

da (1— sin 2 a sin 2 8) 4 

Putting this expression equal to zero, it reduces to : 

(cos 2 a — sin 2 «)(i — sin 2 a sin 2 8) + 4 sin 2 a cos 2 a sin 2 8 = o 

'2 + 3 sin 2 8 N 



sin 4 a -\- snr a 



2 sin 2 8 



2 sin 2 8 



sin a 



^ / (3 sin 2 8 — 2) ± V 9 sin 4 8—4 sin 2 8 + 4 
^ 4 sin 2 8 



which gives two real values of sin a, or four values of a, corre- 
sponding to two maxima and two minima. These values of a will 
be symmetrically grouped about the position « = o (Fig. 21). 
If we know the total moment ; 

of inertia of all masses rotating 
with the follower, we can cal- 
culate the maximum twisting 
moment in the follower due to 
its angular acceleration. As 
an extreme case take 8 = 45 °, 
and take the number of revo- 
lutions per minute at 300. $< 
Then the values of a at which 
the angular acceleration will 
be a maximum will be : 




a = sin -1 ± 



* 



2 ± Vf- 2 



= sin" 1 ± V.7808 



<*]max.= 62° 5' and 242 5', 
a]min.= n7° 55' and 297° 55'- 



30 



MACHINERY OF TRANSMISSION 



Now 



U) l 



2 TtN x 

60 



31.416 radians per second. Therefore substi- 



tuting these values in the expression for angular acceleration, we 
find 

± 776-94 radians per second, per second. 



dt 



Let a cylinder of cast iron 1 ft. in radius, 1 ft. in axial length, 
and of density equal to 450 lbs. per cubic foot, be keyed to the 
follower. If we neglect all other masses, and call the force moment 
Ph, then : 



Ph = I 



dw. 2 
~d~i 



Jf r; = 45o_X Z 
2 64.4 



21-95) 



Ph = 21.95 X 776.94= ± i7>°54> 

or 17,054 lbs. acting on a lever arm of 1 ft. 

In order to transmit a uniform angular velocity ratio by means 
of a universal joint, two joints should be used (Fig. 22). Here the 




Fig. 22 



angle 8 is the same for both, and the axes of the bearings in the 
intermediate shafts' forks are parallel, Then fluctuations in angu- 



AXES OF ROTATION CROSSING 3 1 

lar velocity produced by the first are neutralized by the second. 
Transmission between shafts which are parallel but not in the same 
straight line may be effected by this means also. If the link which 
connects the two joints is made short enough, the two axes or bars 
a'fr' and cd coalesce, and we have the double universal joint. In 
this case, however, we must provide some means for guiding this 
bar in the plane bisecting the angle between the shafts, and there 
is no simple means of accomplishing this. 

4. AXES OF ROTATION CROSSING 

There appears to be no practical means of transmitting rotation 
between crossing shafts by means of a coupling. 



CHAPTER II 

TRANSMISSION OF PURE ROTATION BY MEANS OF 
FRICTION GEARING 

I. AXES OF ROTATION IN ONE AND THE SAME 
STRAIGHT LINE 

A. Direct Connection, Friction Couplings 

The problem of Friction Couplings is purely one of Machine 
Design, and therefore has no place in this discussion. 

2. AXES OF ROTATION PARALLEL 

A. Friction Wheels 

If two bodies have pure rotary motion relatively to a third, their 
centrodes, while rolling upon one another, may be used to transmit 
motion in certain cases. Suppose we have two bodies, b and c 
(Fig. 23), rotating about fixed centres relatively to a. Let the 
angular velocity of b be co : , and that of c be w 2 . Concerning the 
point O hc , or the relative centre of b and c, we know that it must 
lie on the line of centres. Further, from Fig. 15 we know that its 
position on O ah O ac is fixed by the angular velocity ratio, and 
that: 

Obr.Oac _<*>l } 

O bc O ab o> 2 



and since the distance O hc O ac + O hc O ah is constant, the position 
O bc divides a fixed length in the inverse ratio of the angular veloci- 
ties. If the angular velocity ratio varies, O hc will move along the 
line of centres, tracing out a curve with respect to the moving 
plane of c and another with respect to the plane of b. Such 

32 



AXES OF ROTATION PARALLEL 



33 



curves are known as "pitch curves," and will be again considered 
in the case of toothed gearing. They are the centrodes of the 
relative motion, and if cut out of metal and rolled one upon the 




Fig. 23 



other, the original motion would be reproduced. Also power 
could be transmitted provided (1), there is a component of the 
driving force normal to the curves at the point of contact, and in 
the direction of the follower, or (2) if this component be zero, 
that the coefficient of friction be great enough to prevent slipping. 




Fig. 24 



If the curves both make complete and practical revolutions, i.e. 
are closed curves, the first condition is impossible of fulfilment, 
since it necessitates the contact radius of the driver being continu- 

D 



34 



MACHINERY OF TRANSMISSION 



ally on the increase. Hence the only case in which the centrodes 
themselves can be used is the second, and it is applied as follows : 

Let the angular velocity ratio — be constant. Then point O bc 

is fixed on the line of centres, and traces out two circles in the 
planes of c and b, the ratio of whose radii is the inverse ratio of 
the angular velocities. It will be noticed that if the angular veloci- 
ties are in opposite directions, the pitch circles will touch one 
another on their convex or outer sides ; but if w 1 and <o 2 are in the 
same direction, the convex side of the circle having the larger 
angular velocity will touch the concave or inner side of the other. 
(See Fig. 24.) In this case there is no, 
component of driving force in the direction 
of the normal, and the second condition is 
satisfied. 

The difficulty with all such Friction Gears 
is that there is no real tendency of the wheels 
to drive. If a force K (Fig. 25) is to be 
exerted at the circumference and the co- 
efficient of friction is <f>, then the two axles 
must be pressed together with a force Pat 
least equal to 







Fig. 25 






Since P is directly in line with the axles, <£ should be made as 
large as possible in order to diminish P, which causes hurtful fric- 
tion in the bearings. Leather or paper may have <f> as large as .4. 
Another way to diminish friction in the bearings is to have only 
a small component of the force P in their direction. To ac- 
complish this grooved gearing is used. Here a series of circular 
ridges on one cylinder is made to fit a series of grooves on the 
other. The angles of the ridges and grooves are made about 30 
(Fig. 26). Since a point C at the base of a ridge is moving 
slower than a point D at its tip, there will be rubbing between the 
surfaces at both C and D, causing a waste of energy. Hence in 



AXES OF ROTATION PARALLEL 



35 



most cases the surfaces of the ridges and grooves are made curved 

instead of straight, as shown, and therefore touch in but a single 

point. The great disadvantages 

of all friction gearing are, the loss 

of energy at the bearings, due to 

friction there, and the fact that 

the centre distance of the wheels 

must be made adjustable in order 

that wear may be taken up. It is 

impossible to make — absolutely 

constant, owing to the wear and 
slip, and hence it is used in driv- 
ing light-running machinery only, 
except where grooved gearing is 
employed. 

3. AXES OF ROTATION INTER- 

SECTING, AND 

4. AXES OF ROTATION CROSS- 

ING. 



A 




A' 


B 


W^W 


B' 


6 




. 



Fig. 26 



Friction gearing is seldom used in these cases, though any of the 
pitch surfaces fulfilling these requirements (see chapter on Toothed 
Gearing) might be used in a manner similar to the preceding. 






CHAPTER III 

TRANSMISSION OF PURE ROTATION BY MEANS OF BELTS 

AND ROPES 

i. AXES OF ROTATION PARALLEL, PLANES OF PULLEYS 
THE SAME 

A. Ordinary Belting 

In these cases motion is transmitted from a rotating body to an 
intermediate one by pure rolling, and from that similarly to a 
third. The case is a general one, though used extensively in but 
a single form. Let any two bodies b and c have purely rotary 
motion relatively to a third body a. A fourth body d may be 
made to so move that its centrodes, with respect to both b and c, 







Fig. 27 



shall be identical. This may be accomplished by assuming any 
form for the centrodes of b, c, and d, and forcing the three to roll 
together. The only important case is that in which the centrode 
fixed in d is straight, or practically so (Fig. 27) . The centrodes of 
b and c can be cut out of rigid disks, and the straight edge of d will 
roll upon them, transmitting motion by friction, provided no points 
of inflection exist in the outlines of either b or c, and that these 

36 



AXES OF ROTATION PARALLEL 37 

latter do not interfere. The tendency of the centrodes to slip may 
be overcome by forming d of a flexible band, which wraps around 
and is fastened to the outlines of the other curves. This is called 
Transmission through a Wrapping Connector, and power can be 
transmitted in but one direction. The angular velocity ratio will 
evidently be equal to the inverse ratio of the perpendiculars let fall 
from the centres of rotation upon the straight edge of the con- 
nector, or will be equal to the inverse ratio of the segments into 
which the line of the connector cuts the line of centres (produced 
if necessary) . If the motion is to be continuous, i.e. if both b 
and c are to make complete revolutions, the connector cannot be 
fastened to their outlines, but must continuously surround them 
with a belt of constant length. The centrodes of b and c will then 

be circular, — will be constant, and power can be transmitted in 

<*>2 

either direction. 

(a) Stresses in the Belt, and Power Transmitted 

Belts possess the great advantage of lying around a large portion 
of the circumference of the pulleys, and hence, for a given pull P 
between the shafts, create a greater frictional resistance than 
merely P<\>. They also possess the advantage of being elastic, and 
take up by this means any slight variation in the centre distance 
of the shafts, due to wear. The method of finding the total fric- 
tional resistance of a belt at the instant of slipping, in terms of 
the tensions on the tight and slack side, is easily shown by the 
principles of Mechanics to be 

P= 71- T 2 

and T x = T/ a * 

where F is the total frictional resistance, T x the tension on the 
tight side, 7 T 2 .the tension on the slack side, a the angle of the 
pulley covered by the belt, and </> the coefficient of friction 

* Values of e$ a for values of <f> between o and .8 and of a between o and 
360 can be taken from the accompanying Diagram No. 1. 



38 MACHINERY OF TRANSMISSION 

between the belt and pulley. If the force to be transmitted is K, 
then K must be something less than F, or 






K=CF= C(T X - T 2 ) = c(t x -^]= CrJ' 



' e <bi 



\ e!> a J V e*° 

where C is a factor of safety. In most cases of belt transmission 
a = 7r. Now the horse-power transmitted will be 

HP. = v xK = V e* a 

33000 33000 

where v is the velocity of the belt in feet per minute. The only 
variables in the above equation are the horse-power and the 
velocity, hence one varies directly, as the other. 7\ is determined 
as the maximum working tension that the belt will stand. If p is 
the maximum allowable working tension per inch of width of single 
belting, and w is the width of the belt in inches, 

T x _ 3 3000 x H.P. 

/^^ ei>a — 1 



w = 



Fair average values of the constants involved will be about as 
follows: <£ = .25, a = 7r, (Cp) = 55. This last constant includes 
the safety factors of both strength and slip. 

TT 3^000 x H.P. 1062 x H.P. 

Hence, w = — = , 

v x 55 x .545 v 

or as it is usually written 

1000 x H.P. 

w = 

V 

If D is the diameter of the pulley in inches, and N the number 
of revolutions per minute, 

V = ttN — j 
12 

, 3800 x HP. 

and w = - — =z TT 

DxJV 



AXES OF ROTATION PARALLEL 39 

The increase in friction over ordinary friction gears for a given 
pull between shafts, may now be readily computed. Let P be 
the pull between shafts. In the case of the belt we may take 
P=T X + T 2 whena = i8o°. Hence: 

P= T 2 e^ + T 2 = T 2 (e*°- + i), 
T P 

The force transmitted at the circumference of the pulley at the 
instant of slipping is 

K=F=T 1 -T 2 = T 2 (e*« - i ) . 

If <£ = o.28, ^=2.41. 

Hence for the belt, 

^=.413^. 

In the case of friction gears 

X=cf>P=. 2 8P. 

The gain will then be in the ratio of 413 to 280. 

The above formulae can only be considered as an approximation 
to the truth in the case of a rapidly moving belt. In fact they 
give us more the general law of the variation of the quantities 
involved than an exact measure of them. In a rapidly moving 
belt the sudden change in the direction of velocity as the belt 
passes around the pulley creates an additional tension not hitherto 
considered. This may amount to a large proportion of the total 
tension, so that if the belt is tightened up to its safe limit at rest, 
it will be overstrained at high velocity. In fact the velocity may 
become so great as to make the tension due to this centrifugal 
effect equal to the safe tension, so that the belt would not touch 
the pulley at all, if kept within the safe limit. 



40 



MACHINERY OF TRANSMISSION 



Let us examine more closely into this effect. In Fig. 28, the 
normal pressure under an elementary length of the belt due to the 

opposite tensions is 

dN 1 =T- da. 

But this is diminished by 
the centrifugal effect above 
mentioned, so that the true 
normal pressure is 

dN,= T- da - dm -• 
r 

The frictional force is 




Fig. 28 



dF=<f>(Tda-dm- 



But dF is equal to dT, and dm = - • r • da where 8 is the weight of 

the belt per unit of length. But the tension varies directly as 8, 
hence we may write 

8 = nT, 



or 



dF= dT= 4> f Tda - — 2 da\ 



dT 



11 V 



2\ 

in 



Hence, 

and 



IH 1 --)} 



33000 1 e ^ a{1 - z) ) 



The constants of this equation can best be determined experi- 
mentally, or rather should be determined as experience dictates, 



*" Rope Driving," by John J. Flather, p. 115. Wiley and Sons, 1895. 



Lesser Angle of Contact, o<. 



io c 








V 


D 


c 


* 


T 






T 




i 




- 




T 


c 


N 






_ o 










\ 


\ 






H 


ors 


3 PC 


iwe 


r.( 


"or 


sm 


all 


»ull 


eys 


























\ 


X} 












































\ 






\; 


5 


































X 






\ / 


> 




3> 




V 










































; „ 


k* 


r 


£ 


\ 


'<? 










































\- 


























/ 


/ 


X 


X- 


X 


^ 










\ 




\ 






















X 


/- 












X 


1 1 










\ 






































^ 


^ 












\ 




















































\ 


~^ 
















^ 


^ 


^ 


^ 












13 

°- 
































^C 


^ 
















0) 9 


































c 














^ 


J2 


"" 






























^" 


V 
















o 8 
































,>• 


A 














































> 




















o 
















































^ 













> 








f~~ 




j5= 


■^ 


























o 


^r^r 







~^ 


-^ 






r^ 


i"- 




r^r 


__ 







































b 






i 


































^ 






oJL~z 


j£Z 






i 




































*o / 






















































V 


j / 




j . 








































^ 


f- 




*-* 










































6 


S, 










































fc 


22fs* 








































i 


7 




g 










































1 


'U 














j 
















$ 












\\ 




n 


k 


// 














+= 
















-N 


i 










V 



«— CN 



LO l£> 



Velocity of Belt in feet per minute. 



Horse-power Transmitted by Leather Belts per Inch of Width. 

(After Prof. J. J. Flather.) 



AXES OF ROTATION PARALLEL 4 1 

and no attempt should be made to get exact values of cf>, S, and the 
safety factors of slip and strength. In general we can say that as 
the velocity is varied, the horse-power transmitted will follow the 
general law of the equation as given, whose constants are to be 
determined experimentally. Diagram No. 2 shows a series of 
curves drawn by Professor Flather* which takes account of many 
sources of variation in belt transmission. 

To use the diagram, take the speed of the belt at the bottom, 
and follow up to the belt curve required. Then follow horizontally 
to the centre line of the diagram, then parallel to the logarithmic 
curves to the abscissa, showing the angle of the belt at the top of 
the diagram. Going either to the right or left to the margin will 
give the horse-power which one inch width of the belt would trans- 
mit at the given velocity. Suppose we wish to find how wide 
should be a double-laced belt to transmit 30 H.P. at 4100 ft. 
per minute when covering 155 of the pulley. Following from the 
4100 abscissa upwards to the double-laced belt curve, and passing 
horizontally to the centre line, where a equals 180 , we find that if 
covering 180 of the circumference the belt would transmit about 
6.25 H.P. per inch of width. But by following parallel to the 
logarithmic curves to a = 15 5 , it is seen that but 5.5 H.P. will 
be carried per inch of width. Dividing 30 by 5.5 gives for the 
width of belt about 5 J in. 

There are still other causes by which the power will be reduced. 
Air is drawn in by the belt at high speeds, and lessens the normal 
pressure between the belt and the pulley, where the two first come 
into contact. Also, if the diameter of the pulley is comparatively 
small, say less than 20 in., the necessary bending of the belt 
tends to injure it, and hence still less tension can be given. This 
is taken into account in the preceding diagram. If from the point 
last found on the 155 line in the previous example, we follow 
horizontally to one of the diagonal lines springing from the lower 
right-hand corner, and from the intersection with it follow upward 
to the scale at the top, we find the still further decreased horse- 

* Western Electrician, June 12, 1897, 



42 MACHINERY OF TRANSMISSION 

power which each inch width of the belt will transmit. P'or ex- 
ample, if our belt should pass over a 12" pulley, we find the horse- 
power per inch of width to be but 4.45, and that the belt must be 
about 6| in. If necessary, other lines can be interpolated 
between the three already drawn. 

If the angular velocities of the two pulleys are in the same 
direction, the ordinary open belt is used, where the point of inter- 
section of the line of centres with the belt line (produced), lies 
without the space between the centres. If the directions of ro- 
tation of the two are opposite, the crossed belt must be used, and 
the belt cuts the line of centres between the axes. The crossing 
of the two halves of the belt can be accomplished without serious 
interference by giving the belt a half twist. 

The greatest source of loss is by friction at the bearings, due to 
the necessary tension on the belt. This can be roughly calculated 
when the diameters of the pulley and shaft are given," as well as the 
velocity of the belt, the tensions, and the coefficient of friction 
between the journal and its bearing. If «// is the coefficient of 
journal friction, v the velocity of the belt in feet per minute, and 
R and r the radii of the pulley and shaft, then the opposing force 
at the belt due to journal friction will be 

f = (T 1 + r 2) +j i . 

And the power wasted will be 

33000 33000 R 

Hence the tensions should be as small as possible to prevent slip- 
ping, \p should be as small as possible, and r as small, andi? as large 
as possible, for the greatest efficiency. There is also a loss due to 
the creeping of the belt. As the belt arrives at the driving pulley 
in a state of greater tension, and hence of greater stretch than 
when it leaves, there will be a slow creeping. The same is true of 
the driven pulley. This loss does not amount to more than \ of 

1%. 

In running a long horizontal belt it is best to have the tight side 






AXES OF ROTATION PARALLEL 43 

below, as by the greater sag of the slack side the arc of contact on 
both pulleys will be increased. 

A variable angular velocity ratio is not practicable in belted 
transmission. 

(b) Stepped Cones 

When a follower shaft is required to have various speeds ob- 
tained from a constant speed driver, or when in any way the 
angular velocity ratio must be varied with belted connection, we 
must resort to Stepped Cones or an equivalent. This is always 
the case in lathes, milling machines, boring machines, etc. In the 
problem of stepped cones in its simplest form, we are given the 
distance between shafts, the diameters of the steps on one cone, 
and diameter of one step on the other cone ; and we are required 
to find (1) the length of belt that will fit the given pair of steps, 
and (2) what must be the rest of the steps so that this same belt 
will fit them all with equal tension. 

If the belt is crossed, the problem is a simple one. In Fig. 29 

let / be the half length of the belt, a the angle between the belt 

and the line of centres, and R and r the radii of any pair of steps. 

Then 

d since = R -\- r I. 

Also, l=dcosa + - R + aR + - r + ar, 

2 2 

= dcosa + -(R + r) + a(R + r) . . II. 

2 

The length of belt which will fit the given pair of steps is now 
immediately known, for a is calculated from No. I, and / from 
No. II. Hence / becomes a known quantity. If now we sub- 
stitute for cos a and a in No. II their values as deduced from 
No. I, we get 

/= Vd 2 ~(R + rf + {R + r )fc + sin" 1 *±iY III. 

In this, if the radius of any step of one cone, such as R, be given, 
the only unknown quantity is r, which would be the corresponding 
radius of the step on the other shaft, around which the already 



44 



MACHINERY OF TRANSMISSION 



calculated belt length would fit. But equation No. Ill is transcen- 
dental, and of such a form that a direct solution is impossible. 
However, if we substitute in No. II the value of r or R deduced 
from No. I, both R and r eliminate from the equation. But R 
and r are variables. Hence «, the only other unknown quantity, 




Fig. 29 



must be a constant, as neither of the variables can be expressed 
in terms of it. If a is constant, R + r is constant, or 

R\ + n = R 2 -f- r 2 = R 3 + r 3 = • • • and R x — R 2 = r 2 — r 1? 
R 2 -R 3 = r 3 -r 2 , etc., 

and the steps of the two cones are equal. This is readily seen to 
be true in equation No. II, for if R + r is constant, / is con- 
stant, a necessary condition. 

The case of the open belt is more complex, and an exact analytic 
solution has not yet been arrived at. In this case we have from 
Fig. 30, 



and 



d sin « = R — r 
/= d cos a 



IV, 



- R + aR ^- - r— ar 

2 2 



= d cosa + - (R + r) + a (R - r) 

2 



Substituting for « and cos a as before, we get 



/= Vd 2 - (R - rf + - (R + r) + (i? - r) sin" 1 ^— ^ VI, 



AXES OF ROTATION PARALLEL 



45 



which is transcendental also, and not capable of algebraic solution. 
Furthermore, if we substitute for r or R in No. V their values as 
deduced from No. IV, we get 

R = — (« sin a + cos«) + - sin a, 

7T 7T 2 



— (« sin« -h cos u) — — sin a, 



which differ from one another only in the sign of the last term. 
But a is an unknown variable, and, as can readily be seen by 
inspection of the figure, cannot be a constant. Hence algebraic 
solution of the open belt is impossible. 




Fig. 30 

Reuleaux* has deduced an exact graphical solution of the above 
equations, but it involves the tracing of certain higher plane curves, 
and is therefore not practical. His extremely beautiful and in- 
genious solution is as follows: in Fig. 31, draw BC and AD 
parallel, and at a distance apart equal to the shaft distance d. 
Draw AB at right angles to AD. With A as a centre and with a 
radius equal to AB draw the quadrant BE, which quadrant will 
contain all possible values of «. Draw the involute EF of the 
quadrant BE, with cusp at E. Let EAP be any arbitrary value 



* "Constructor," p. 1S9. 



4 6 



MACHINERY OF TRANSMISSION 



of a. Draw the tangent at P, cutting the involute at N. Then 
PN=PE. Draw PM perpendicular to AD, NK perpendicular 
to PM, and Q Y perpendicular to AD, passing through N. Then 

A Q = AM+ MQ = d (a sin a + cos a) . 

Now make B C=7rd—7r(AB) , and complete the rectangle ABCD. 
Draw the diagonal BD. This diagonal must make with BC an 
angle whose cotangent is tt, or 17 39' 19''. Then 



YG = - (a sin a + cos a), 




31 E Q H 

Fig. 31 



which is the middle term of the equations for R and r. Again, if 
we lay off BL = I, the half length of the belt, which can be calcu- 
lated for the given pair of steps, then 

LS=YJ=-, 

TV 

which is the first term in the equations, and then we have 
JG = YJ- YG= (a sin a + cos a). 

7T 7T 

Finally, if we draw a semicircle on \AB as a diameter, then 



^r=-sin a. 
2 



AXES OF ROTATION PARALLEL 47 

This we may add and subtract from/^ to get the values of R and 

r corresponding to any chosen value of a. Lay off AT upwards 

from G and downwards from G, determining the points £7 and V. 

Then 

JU=R 

and JV= r. 

Now similar points may be obtained for other values of a, and thus 
a smooth curve HVXUF is drawn, tangent to WE at X, and sym- 
metrical as to ordinates about BD as an axis. The assumed pair 
of radii by which / was calculated must fit the curve, and thus may 
check the work. But we may even find the value of / from the 
diagram without calculation, for the curve HVXUF may be drawn 
without reference toy at all. Then if we assume our first pair of 
steps, we will know their difference, which is equal to UV. So we 
merely need to try along the curve till we find an intercept UV 
equal to (R — r), then lay off UJ equal to R, or VJ equal to r, 
and the pointy becomes immediately fixed. 

Hence Reuleaux's solution is exact and complete, but it is labo- 
rious, and involves the plotting of the involute EF, and the curve 
HVXUF. 

Lately Professor W. K. Palmer* has proposed a modification of 
Reuleaux's method. He introduces a negligible approximation, 
but the method is eminently practical. Instead of plotting the 
curve HVXUF as double ordinates QU and QV laid off along 
the same line, and therefore in the same direction, he lays off 
QU as an abscissa and QV as an ordinate with reference to A as 
an origin of rectangular coordinates. It will be noticed on our 
old diagram that as we pass along the curve from H towards V, 
the values of R and r become equal at X. When we pass beyond 
this point, the quantities R and r exchange places, and all those 
values formerly given to R must now be given to r, and vice versa. 
Also that the maximum value that can be given to R is when /= icd. 

* "The Designing of Cone Pulleys," by Walter K. Palmer, Lawrence. 
Kansas. 



4 8 



MACHINERY OF TRANSMISSION 



Then R = d and r = o, or r = d and R — o. On our new diagram 
then (Fig. 32), if BFE be our new curve plotted in rectangular 
coordinates, it must be symmetrical with respect to a 45 ° line AF 
through the origin A, Furthermore, it must pass through the points 
B and E of the old diagram. Now in Fig. 31. 



R=JU= QU- 
r=JV= QV- 






Hence to correctly represent the values of R and r on our new 
diagram, we must subtract the constant length QJ from both the 




Fig. 32 

ordinates and abscissae of the new curve. This can be done by 
shifting the origin, from which we measure our step radii, along 
the 45 line from A to A'. Now 

p = WP= YP - YW 9 

r=XP = MP-MX, 

and XM= YW. Hence 

R = A'X and r=XP. 

The point A' is determined by the length of the belt, and by the 
diagonal BD exactly as/ was determined in Fig. 31. But we may 
even do away with the calculation of /, and find it directly from the 
diagram, for any given pair of steps will give the point A\ Make 
AI=R and IT= r for the assumed pair of steps. Draw AT. 



AXES OF ROTATION PARALLEL 49 

Now shift the triangle AIT, so that its angle at A remains on the 
45 line, and its sides AI and IT remain parallel to the coordinate 
axes, till the angle at T touches the curve at P. This can be done 
by drawing TP parallel to AF, and PA' parallel to TA. Then A' 
is the new origin from which BL = I can be obtained. All other 
pairs of radii such as P 1 = A'G, ?\= GH, etc., can be drawn. 
Palmer finds that the arc BFE can be very exactly approximated 
by the arc of a circle whose centre lies on the 45 line produced 
below A a distance equal to the diagonal of a square whose side is 
one-tenth of d. This removes all necessity of plotting curves of any 
sort, and leads us to a direct and practical result. 

It is of interest to notice that in the case of crossed belts the 
circular arc BFE in Palmer's diagram becomes a straight line BE, 
since the coordinates of points along such a line measured from any 
set of axes parallel to AD and with origins on AF will have a con- 
stant sum. Hence R + r = constant, the necessary condition for 
crossed belts. Otherwise the construction is the same as for the 
open belt. 

When the distance between shafts is great as compared with 
the radii of the pulleys, we may consider a equal to zero in the 
equation 

/= d cos a + - (R + r) + a(R- r), 

and any slight error due to this assumption will be made up by the 
elasticity of the belt. In this case our equation reduces to 

Here, as in the case of crossed belts, if we make R + r constant, the 
equation is satisfied, or the steps on the two cones will be equal. 

Professor John E. Sweet* has employed the following graphical 
method of designing the steps of cones, where the distance be- 
tween shafts is great enough to allow 7 equal steps. It is as follows : 

* American Artisan, February, 1 874; American Machinist, October 13, 
1898, p. 757. 
£ 



50 



MACHINERY OF TRANSMISSION 



Let A and B (Fig. $$) be the centres of the pulleys, B being the 
countershaft, which runs at a constant angular velocity, wj. Since 
the speeds vary inversely as the pulley diameters, we lay off the 




Fig. 33 

velocity of B from the centre of A, and in any direction as AX. 
Similarly we lay off the smallest velocity of the driven machine 
(back gear out) from B parallel to AX and equal to BZ. Produce 
AB and draw XZD. Now 

Ang. vel. A : Ang. vel. B : : BZ: AX : : BD : AD. 

Let BF be the radius of any step of the cone, assumed to begin 
with. Draw a line from D tangent to circle BF, and produce 
toward A. Draw a circle about A as a centre, and tangent to 
DF at G. Then 

GA:FB::AD: BD : : Ang. vel. B : Ang. vel. A. 

Now lay off BY equal to the next speed required of the machine 
A. Draw XYE. Take C at the middle point of AB and erect 
a perpendicular CQ intersecting GF at P, then P is the middle 
point of GF (nearly) . Connect EP and produce. Draw circles 
about A and B tangent to this line, and these will be the next pair 
of steps. It will be noticed that since P is the middle point 
of GF, the steps will be equal, or the same belt will fit all steps. 
Modifications have been proposed for the above to suit cases 
where the shaft distance is not great enough to allow equal steps.* 



* C. A. Smith, American Machinist, February 25, 1882; John Coffin, 
American Machinist, April I, 1882. 



AXES OF ROTATION PARALLEL 



51 



The angular velocity ratio of a pair of stepped cones should 
form a geometrical progression, and if the driven machine is 
back geared the back gearing should carry the geometrical pro- 
gression on. Let us suppose that we have a three-stepped cone, 
with back gear, which runs at ten revolutions per minute with the 
back gear in and the belt on the step of largest radius. Further- 
more, suppose that each succeeding step increases the speed 50%. 
Then we would have 

fStep(i), 



10. 



Back gear in -I Step (2), 1.50 x 10.00 =15. 
I Step (3), 1.50 x 15.00 = 22.5 



fStep(i), 1.50x22.50 = 33.75. 

Back gear out j Step (2), 1.50 x 33.75 = 50.52. 

I Step (3), 1.50 x 50.62 = 75.94. 



The ratio of the back gearing would be 
P 



IO I , , s 

= - (nearly). 

33-75 3 



If these speeds are laid off as ordinates equally spaced along a 
horizontal line, we get a smooth curve (Fig. 34). If, however, we 
find by trying the differ- 
ent speeds of a cone- 
driven machine that the 
curve is irregular, then 
the cones are badly de- 
signed. This method 
of testing the cones of 
a lathe has been sug- 
gested" by Professor 
Sweet. He also uses 




Fig. 34 



the following method of graphically computing the geometrical 
progression of speeds : lay off PO (Fig. 35) equal to 100 units, 
and OA equal to the number representing the percentage of 
increase in speed. In the preceding case OA = 50. Erect a 



52 



MACHINERY OF TRANSMISSION 



perpendicular at O and another at A. Lay off Oa = 10, or the 
slowest speed of the cone with back gear in. Draw PaB ; then 
AB — 15, the speed of the second step. Square back from B to 
b, so that Ob equals AB, and draw PbC. Then AC= 22.5, the 
third speed, and so on. 




Fig. 35 



Frequently the extreme speeds of the driven machine are given, 
and we are required to divide up the interval according to a 
geometrical progression. If a is the slowest speed or the first 
term of the series, and x the constant multiplier, then our series is 

a, ax, ax 2 , ax 3 , ax*, etc. 

The nth. term will be ax 71-1 = b. In our problem we are given 
a, b, and n to find x. 



x 



•% 



and the ratio of the back gear will be 



In the preceding problem a = 10, b = 75.94, n = 6. 



75-94 






3-375 



Sweet's method of constructing the geometrical progression of 
speeds can be combined in a very elegant manner with Palmer's 



AXES OF ROTATION PARALLEL 



53 



method of designing the cones, as the following problem will 
show: let the distance between the shafts of a foot lathe be 
30". Let the radius of the largest step on the driving wheel be 
io|", and of the smallest step on the headstock be 1,5". The 
maximum angular velocity ratio will then be y = 7 (back gear 
out). Furthermore, let the minimum angular velocity ratio be 
y= 1 (back gear in), and let there be three steps to the cones, so 
that n — 6. Then 

£ = JJ>max. _ 7 

a 



max. 

7min. ~" I ' 



The ratio of increase in speed is 

x=\ r j= i-475 8 , 
and the ratio of the back gear is 



P = 



i-475 8 



3.2141 



- (nearly), 



Lay off AB equal to 30" (Fig. 36), and AK and KO each 3", 
determining the centre O. Draw the arc BE about O. Lay off 




Fig. 36 



from origin A, R±= 10.5", and r, = 1.5". Shift origin to A 1 , and 
produce A'S. Make A'M equal to 100 units of any convenient 



r x 


= i". 5 p 


r 2 


= 2.14. 


H 


= 2.96. 



54 MACHINERY OF TRANSMISSION 

scale, and MN equal to 47.58 units. Erect perpendiculars at 
M and N. Produce A'P to Q. Square back to U and draw 
AU, etc. Measurement of a large scale drawing gives us the 
following results : 

^=io". 5 o 

d= 30" \ R 2 = 10.10 
LiP 8 = 9.50 
The half-length of belt is 

BL = IA = 48".04, or 8' Jg." for the whole length. 
The ratio of the back gears is \. 

B. Wire Ropes 

For transmitting power over great distances, wire rope is used 
in place of belting. In this case the rope runs in grooved wheels 
of large diameter, known as sheaves. The bottoms of the grooves 
are often lined with wood or rubber to increase the coefficient of 
friction. The diameter of the sheaves being great compared with 
the diameter of the journal, the loss of power in the boxes is 
reduced to a minimum. The rope is supported at intervals by 
smaller pulleys known as idlers. 

In the rope transmission problem, the horse-power to be trans- 
mitted, the velocity of the rope, and the distance between idlers 
are given. The tensions on the tight and slack side, which are 
necessary to transmit the required horse-power at the given veloc- 
ity at the instant of slipping, can be calculated approximately from 

H.P. x 33000 



T,= 



— 1 



and j 



^1 



Then if the rope is at rest, the tensions on the two sides will be 
equal, and will be approximately 



AXES OF ROTATION PARALLEL 



55 



Generally T x is taken equal to 2 7^, which gives with a = tt the co- 
efficient of friction <£ as only 0.22. This then allows an ample 
factor of safety ; in other words it will include the factor C, and the 
mean tension will be T= J 7j. 

Now to find what size of rope will bear the maximum tension 
T lt we must look in some table of wire rope sizes and strengths, 
select that size which will bear T x as a working stress, and from the 
same table find the weight per foot of such a rope. The rope is 
then drawn up to the required tension T, and spliced in place. 

The best way to measure the tension in such a rope is to meas- 
ure the sag or deflection at the middle point of the span. The 
calculation of this sag we will now take up. 

{a) Horizontal Transmission 

Let POQ (Fig. 37) be a suspended rope, whose points of sus- 
pension P and Q are in a horizontal plane. Then O the lowest 
point will also be the middle point. The coordinates of P are x 




Fig. 37 



and y when referred to an origin at O, with axis of X tangent to 
the rope {i.e. horizontal). The tension at .Pcan be resolved into 
its vertical and horizontal components, which are 



V= Tsm a, 
H= Tcos a, 



V , 
— = tan a 
H 



ay 
dx 



56 



MACHINERY OF TRANSMISSION 



TABLE 

Standard Iron and Steel Hoisting Ropes, John A. Roebling's 
Sons Company 

Wire Ropes of 19 Wires to the Strand 









T3 

a 
3 



Iron 


Cast Steel 




a 


c 


ft-* 


*o 


«j 


01 


g-s 


l*H 








a 








c cr 


n 








c §• 




1 

3 

a 
D 


22 
1 


U 

c 

"I 



,o 

u 

Pa 

X! 
be 


a 
_c 
'« . 

bJD 
c 

Si ° 


M O 

.5 8 


u a. 

6 5 c 
§ « Si 


6*5 

<i2 

I'i 

.§s 

ex: 


.5 
c 
'« • 

bfl 

•5 8 

« O 


« «; 
°£ 

■ bf 

.5 8 

^0 


8 p. 

c 2 

3 « « 


« . 

6S 

'-5 a 

IS 

.5-5 


H 


Q 


O 


£ 


PQ 


£ 


O 


S 


pq 


£ 


U 


3 


I 


2| 


6f 


8.00 


74 


15 


14 


13 


155 


31 


— 


81 


2 


2 


6 


6.3O 


OS 


1.3 


13 


12 


125 


25 


— 


8 


3 


T 3 
*4 


52- 


5-25 


54 


II 


12 


IO 


106 


21 


— 


71 


4 


*8 


5 


4.IO 


44 


9 


II 


81 


86 


17 


J 5 


H 


5 


,1 
J 2 


4i 


3.65 


39 


8 


IO 


7l 


77 


15 


14 


5* 


51 


if 


4t 


3.00 


33 


6* 


9l 


7 


63 


12 


13 


Si 


6 


I* 


4 


2.50 


27 


5l 


81 


61 


52 


IO 


12 


S 


7 


I* 


3l 


2.00 


20 


4 


7l 


6 


42 


8 


11 


4+ 


8 


I 


3* 


I,S8 


16 


3 


61 


5l 


33 


6 


9* 


4 


9 


^ 


2f 


I.20 


11.50 


*1 


Si 


4* 


25 


S 


81 


31 


10 


3 
4 


*1 


O.88 


8.64 


*t 


4| 


4 


18 


31 


7 


3 


101 


5 
8 


2 


0.60 


5-1.3 


t1 

*4 


3l 


3 2 


12 


2i 


5i 


*1 


w>4 


T 9 6 


I* 


O.44 


4.27 


* 


02 


2| 


9 


,1 


5 


If 


1 of 


1 
2 


ii 


o-35 


3.48 


1 


3 


2i 


7 


I 


4l 


II 


10 a 


T V 


1* 


0.29 


3.00 


* 


2* 


2 


52" 


¥ 


4f 


li 


IO| 


8" 


T l 

x 4" 


0.26 


2.50 


1 


2* 


I* 


4l 


1 
2 


3i 





Wire Ropes of 7 Wires to the Strand 



II 


II 


4l 


3-37 


36 


9 


IO 


13 


62 


13 


13 


81 


12 


I 8" 


41 


2.77 


30 


71 


9 


12 


52 


10 


12 


8 


13 


*1 


3t 


2.28 


25 


6i 


«1 


1 of 


44 


9 


II 


71 


14 


II 


3t 


1.82 


20 


5 


71 


91 


36 


71 


IO 


61 


15 


I 


3 


1.50 


16 


4 


61 


81 


30 


6 


9 


51 


16 


7 
8 


2f 


1. 12 


12.3 


3 


51 


71 


22 


4l 


8 


S 


17 


3 

4 


2f 


0.88 


8.8 


2i 


41 


<* 


17 


31 


7 


41 


18 


1 1 
16 


21 


0.70 


7.6 


2 


41 


6 


H 


3 


6 


4 


19 


5 


T 7 
*8 


o-57 


5.8 


T l 

X 2 


4 


51 


11 


2i 


51 


31 


20 


_9_ 
16 


T 5 
*8 


0.41 


4.1 


I 


31 


41 


8 


,3 
x 4 


41 


3 


21 


1 


T3 
*8 


0.31 


2.83 


£ 


2| 


4 


6 


,1 


4 


2! 


22 


tV 


T l 
I 4 


0.23 


2.13 


1 


2! 


31 


4 2 - 


*1 


31 


2l 


23 


1 


tl 


0.19 


1.65 




2i 


2| 


4 




34 


2 


24 


5 
T6 




0.17 


1.38 


— 


2 


21 


3 


I 


2} 


,3 

Z 4" 


25 


9 
32 


¥ 


0.125 


1.03 


— 


t3 
*4" 


2i 


2 


1 


H 


*2 



AXES OF ROTATION PARALLEL 57 

Now the horizontal component of the tension is necessarily con- 
stant in magnitude, and if we consider the weight of the rope as 
uniformly distributed along its horizontal projection PQ, then 

approximately 

V = wx 

where w is the weight of the rope per running foot. So our 

equation is 

dy _wx 

~dx~~H' 

w x 2 

The equation represents a parabola with origin at the vertex. 
From the triangle of forces at P 

T 2 = J? 2 + V 2 , 



So our equation becomes 

w s 2 

y = 



j T 2 W 2 S 2 8 

^ 4 



where y is the sag or deflection at the middle, and s= 2 x is 

ID'S 

the span between idlers. In most cases the term is insig- 

4 
nificant compared with T' 2 , and can be neglected, hence the 

formula simplifies to the extent 

y 8T 
As an example of the use of the formulas, take the following : 
In a horizontal transmission let the span be 250 ft., the power 
to be transmitted is 100 H.P., and the velocity of the rope is 
4000 ft. per minute. We will assume 7^ = 2 r 2 . 

^ H.P. x 33000 , „ 

2" v 

r 2 =~ ^825 lbs. 
2 J 

T=^±B= 1237 lbs. 



58 



MACHINERY OF TRANSMISSION 



By reference to the table it is seen that the diameter of a rope, 
which is to stand a working load of 1650 lbs., is T 9 g in., and that 
the weight of such a rope per foot is 0.44 lbs. The deflection at 
rest is 

ws 2 



The deflection at full load is 



O.44 x 62,500 

5 = 2.5 It. 

8 x 1237 



and 



IDS 

yi = ——=2.oS ft. on the tight side, 
8 J 1 

72 = ^-^7 = 4.16 ft. on the slack side. 
8 T 2 



(b) Inclined Transmission 

In the case where PQ is not horizontal, as in Fig. 38, the 
method is slightly more complicated. As before, we consider 




Fig. 38 



the weight of the rope uniformly distributed along the chord PQ ; 
hence, if this distance be called r, the total weight of the rope 



AXES OF ROTATION PARALLEL 59 

will be w ' r, where w is the weight per running foot as given 
from the tables. But the weight will be distributed uniformly, 
though at a different rate, along the horizontal projection s. Let 
the weight per foot distributed in this way be w Q . Then 

r a/c 2 + s 2 
s s 

c being the difference in level between the points P and Q. 
Taking as before the origin at the vertex of the parabola, and 
calling the coordinates of Q, a, and /?, 

T' 2 = & 2 + w 2 (a + s) 2 . ... 1 
T» 2 = H 2 + w Q 2 a 2 . . . - . . . 2 

* = f£ 3 

2 Jo. 

2 12 

In these four equations, s and c are known from the conditions 
of transmission. One of the tensions, viz., T", is known from the 
horse-power conditions. This leaves four unknown quantities, 
a, (3, T', and H, to be found from the four equations. They can 
be combined with the following result. As equations 2, 3, and 4 
contain but three unknowns, we combine them first. Taking 
3 and 4 : 

WqCi 2 w (a + s) 2 w s(2a + s) 

Yh +c=z 2H > H= Yc * * 5 

Combining 5 with 2 : 

W 3 ± ^4 T " 2 (c 2 + s 2 ) - wfs\ 6 

H then becomes known from 5, pi from 3, and T 1 from 1. 

For example, in an inclined transmission, let the horizontal 
distance between points of support be, s= 250 ft., and the ver- 



6o 



MACHINERY OF TRANSMISSION 



tical distance between the same points be, ^=150 ft. Eighty 
horse-power are to be transmitted at 4000 ft. per minute velocity 
of rope. Then (Fig. 39) 

80 x 33000 =qo 
K4000) * 

7- = ^00 = 650 lbs. 



T= 1300 + 650 ^ 



975 lbs. 




Fig. 39 
Select as before a j^-rope weighing 0.44 lbs., per foot. 



Wc 2 + s 2 V25o 2 +i5o 2 „ 

w = w = .44 — = 0.5 13 lbs. 

s 250 

Deflections at full load : 



«! = 



- .513 x 25o 3 ± 150V4 x i3Qo 2 (i5o 2 + 25o 2 )-.5i3 2 X25o 1 
2 x .5i3(25o 2 + 150 2 ) 



Since in our case s is positive, and is measured in the same 
direction as a, we take the positive value. 

«!= 1207.5 ft - (or - ^S* 1 ^ ft -)- 



AXES OF ROTATION PARALLEL 6 1 



ai tt •S I 3 X 2 5°( 2 X 1207. c + 250) 

Also H 1 =-^~^ ^— - / -^- J — ^-^=1139.6 lbs. 

2 x 150 oy 

5 i3Xi20 7 5 2 = t 

^ 2x1139.6 "> ^ 



?!' = Vii39- 62 + .5132(1207.5 + 250)2= 1363. 1 lbs. 

The deflection at the middle is taken as the difference in level 
between the middle point of the chord r (see Fig. 38), and the 
point of the rope vertically below it. Calling the abscissae of 
these points Y± and Y±, and the deflection y lf 

y^Y.-Y'. 

F 1 = ft + ^ = 403.3ft. 

*i'= V '' = 399-74 ft- 

2 /Z x 

^=3.52 ft. 

Similarly for the slack side, 

7]," = 650 lbs. 
« 2 = 551.6 ft. (or -73542 ft). 
#2 = 578.63 lbs. 

&= 134.9 ft. 

TJ = 709.91 lbs. 

Pa= 2O9.9O ft. 

Y 2 ' = 202.99 ft- 
y 2 = 6.91 ft. 

Similar figures might be obtained for the rope at rest. 

The material of the rope is subjected to two tensions, — the 
working tension, and the tension due to bending around sheaves. 
The working tension may therefore be greater as the bending 
tension is less, or as the diameter of the sheaves becomes larger. 
The size of sheave which will keep the bending tension down to 



62 



MACHINERY OF TRANSMISSION 



a low limit is as follows (expressed as a multiple of the rope 

diameter) : 

Rope of 7-wire strands, Ratio =150. 

Rope of 12-wire strands, Ratio=ii9, 

Rope of 19-wire strands, Ratio = 90. 

When the velocity of the rope is high, the effect of centrifugal 
force will be felt exactly as in the case of belts. If the span is less 
than 60 ft., the transmission cannot be effected economically. 
For distances up to 2000 ft., such transmission is very efficient. 



2. AXES OF ROTATION PARALLEL, PLANES OF PULLEYS 
DIFFERENT 

In belting up shafts where the medial planes of the pulleys are 
not identical, care has to be taken that the belt does not run off. 




Fig. 40 




Fig. 41 



If the belt approaches the pulley at an angle other than 90 with 
the axis, it will fall to one side because the inclination causes it to 



AXES OF ROTATION PARALLEL 63 

describe a helical path. On the other hand, in leading the belt off, 
the receding part may be given an angle of departure as great as 
20 without causing it to fall off, because in this case the actual 
friction between the belt and pulley must be overcome. In every 
instance the point at which the belt is delivered from one pulley 
must lie in the medial plane of the next pulley. This is the only 
condition to be fulfilled, provided the angles of departure do not 
exceed 20 . By properly applying the above condition, and by 




the use of idlers or guide pulleys, which must themselves comply 
with the condition, parallel shafts can be belted up where the 
planes of the pulleys are different. Fig. 40 shows one method of 
doing this. The belt leaves pulley A at an angle of departure 8, 
but the guide pulley c delivers it in the medial plane of B. The 
guide pulley d similarly handles the part receding from B. It will 
be noticed that the guide pulleys will have their planes determined 
by the belt lines, and that the belt must be given a quarter twist in 



64 MACHINERY OF TRANSMISSION 

passing around them. The above arrangement can be run in but 
one direction as shown by the arrows, but if the shafts are to turn 
in opposite directions, the arrangement of Fig. 41 can be used, 
and the belt run either way, when the diameters of both guide 
pulleys are equal to the distance between the planes of A and B. 
It is not possible in any case to dispense with either of the guide 
pulleys. 

3. AXES OF ROTATION INTERSECTING 

Fig. 42 shows a method of belting a pair of shafts whose axes 
intersect at O. The belt leaves the pulley B at an angle of depar- 
ture, but the guide pulley c delivers it to B in the medial plane of 




Fig. 43 

the latter. The whole method is practically identical with the 
preceding. If c and d are so placed that the angles of departure 
of the belt are zero, the apparatus will run in either direction. If 
the medial plane of B is made to include the point at which the 



AXES OF ROTATION CROSSING 



65 



belt is delivered. from A, the guide pulley c can be dispensed with, 
but in this case the angle between the shafts cannot be greater 
than 20 . In no case can both guide pulleys be dispensed with 

(Fig. 43 )• 

4. AXES OF ROTATION CROSSING 

In this case the method of procedure is the same as in the 
former. In one particular instance both the guide pulleys can be 
omitted, as in Fig. 44. Here the radius of pulley A is equal to 




Fig. 44 



the distance from its shaft to the plane of B, and the radius of B 
is similarly proportioned with respect to A. The belt will then 
run only in the direction shown by the arrows. 



CHAPTER IV 

TRANSMISSION OF PURE ROTATION BY MEANS OF 
TOOTHED GEARING 

i. GENERAL CONSIDERATIONS 

When a constant angular volocity ratio is required to be main- 
tained at every instant of time, toothed wheel transmission is 
usually resorted to. In this form certain projections upon and 
depressions within the rotating bodies are made to mesh, so that 
they are driven by direct contact. Any sort of projections and 
depressions will cause one wheel to follow the other, and will keep 
up an average constant velocity ratio, provided they neither inter- 
fere nor fail to engage at any time. But if an absolutely Constant 
angular velocity ratio is required at every instant, the outlines of 
the teeth must be formed according to certain laws. 

In all cases of toothed gearing, the teeth are considered as fixed 
upon certain imaginary surfaces known as Pitch Surfaces. These 
are of such a form that any pair will be tangent along a straight 
line, and that when rotated about fixed axes, with equal component 
velocities at right angles to the line of contact, the required motion 
of the gears will be reproduced. In general such motion will be 
one of pure rolling normal to the line of tangency, and of sliding 
along it. 

2. AXES OF ROTATION PARALLEL, SPUR GEARS 

A. Laws of Action 

In this, the most important case, the pitch surfaces are cylin- 
drical, and tangent along an element which lies in the plane of the 
axes. Here, since the resultant velocity of a point on the surface 

66 



AXES OF ROTATION PARALLEL 



6/ 



is normal to an element, the motion will be one of pure rolling, 
with no sliding along the element. The sections of the cylinders 
at right angles to the axes being parallel, the relative motion is 
evidently uniplanar ; hence we need consider the motion in but a 
single plane. The curves of intersection of the pitch surfaces by 
the normal planes, are known as Pitch Curves, which must have a 
point of tangency on the line of centres. 

Profiles. — Two curves are said to be profiles when during any 
relative motion they remain tangent. The case of two free profiles 
is one of little practical use, but 
when we impose the further con- 
dition that they rotate about fixed 
axes, especially when these are par- 
allel, it becomes one of great im- 
portance. 

Let xx and yy (Fig. 45) be two 
profiles, which rotate about centres 
A and B, at angular velocities o^ 
and o> 2 . Let x be the driver and 
y the follower. P is the point of 
contact of the profiles, and PA = r x 
and PB = r 2 are the contact radii. 
Consider P to be a point of x. Its 
velocity z\ will be at right angles to 
r x and equal to r^. The compo- 
nent of i\ along the tangent to the 
profiles at P will be i\m x cos a, and 
the component along the normal 
will be 7\w x sin a. Consider now 
P as .a point of y. Its velocity v 2 is r 2 o> 2 , and its component 
resolved as before are r 2 w 2 cos j3, and r 2 aj 2 sin (3. If the curves 
are to remain continually tangent (i.e. are to be profiles), 




Fig. 45 



r^tox sin a = r 2 co 2 sin (3, 

w i _ r i sin /? 
w 2 r x sin a 



68 MACHINERY OF TRANSMISSION 

otherwise they would either leave contact or cut into one another. 
Now from the centres A and B drop perpendiculars AG and BH 
upon the normal. Then, 

AG = r x sin a, 

BH=r 2 sin (3, 

oi, BH 



0> 2 


AG 


But if the normal cuts the line 


of centres at I, 


BH 


BI 






AG 


AI 


0)i = 


.BI 


a) 2 


" AI 



Thus in direct contact transmission, the angular velocity ratio is 
inversely proportional to the segments into which the line of cen- 
tres is divided by the common normal to the profiles. 

The above proposition can be proved by the instantaneous cen- 
tre relation also. The bodies x and y are rotating about fixed 
centres A and B. The instantaneous centre of x referred to y 
must lie somewhere on the line connecting these centres. Now 
the point P, when considered a point of x, must be moving in the 
direction of the common tangent relatively to y, from the profile 
condition. Hence O xv must lie somewhere on the line drawn at 
right angles to this direction, or somewhere along the common nor- 
mal. It therefore lies at the intersection of the normal with the 
line of centres. But from Fig. 15 we know that 



Ml __BO 






o> 2 A O xy 
Hence the proposition is established. 

If the ratio — is to remain constant we must have 

oj 2 

BI 

— = const. 

AI 

But BI-\-AI= const. 



AXES OF ROTATION PARALLEL 69 

Hence / must be fixed on the line of centres. This important 
proposition can be summed up as follows : when profiles are to 
transmit a constant angular velocity ratio, they must be so formed 
that their common normal continually passes through a fixed point 
on the line of centres, which point divides the centre distance in 
the inverse ratio of the angular velocities. Profiles so related are 
known as Conjugate Profiles. 

The velocity of sliding between the curves will be the difference 
of their velocities along the tangent, or 

v s — r 2 (o 2 cos (3 — r 1 o) 1 cos a. 

If the velocity of sliding is zero, there will be pure rolling between 
the profiles, which then become centrodes. In that case 

r^i cos a = r 2 <i> 2 cos /3. 
But also 

r^ sin a = r 2 w 2 sin /?. 

Both these conditions can be satisfied only when v x — v 2 both in 
magnitude and direction. But zf 1 and v 2 are at right angles to r x 
and r 2 , hence for pure rolling, P must lie on AB. This is easily 
seen to be true, since for pure rolling, P must be O xy . 

Conditions to be Fulfilled. — The conditions to be realized if 
possible are : 

1. That the angular velocity ratio be constant. 

2. That there should be positive driving, and that, therefore, 
there should be a component of driving force in the direction of 
the normal, and on that side of the tangent occupied by the 
follower. From this it follows that the contact radius of the driver 
must be on the increase. 

3. The velocity of sliding should be zero if possible. 

All three of these conditions cannot be fulfilled, for if (3) be 
true, P must continually lie on the line of centres, and if in 
addition (2) be true, the common tangent at P must cut the line 
of centres at some angle other than 90 . Hence the point of con- 
tact cannot remain fixed on the line of centres, but must move, 
thus giving a variable angular velocity ratio. Similarly if any two 



;o 



MACHINERY OF TRANSMISSION 



of the three conditions be fulfilled, the third will fail. This gives 
rise to three general forms of such driving : 

When (i) and (2) hold, and (3) is abandoned, we have the 
ordinary toothed gearing. 

When (1) and (3) hold, we have the case of friction wheels. 

When (2) and (3) hold, we have some variable angular velocity 
ratio, such as may be produced through limited angles by equal 
ellipses or logarithmic spirals. 

The second case we have already considered, and the third is of 
little value, though the elliptic form as applied to a pitch curve will 
be treated in its proper place, but the first is of great importance, 
and we now proceed to take up its study. 

Curves of Action. — Let aa and bb (Fig. 46) be two pitch 

curves rotating about A and B. If there is to be pure rolling at /, 

*>i BI Tr , 

— = — - • If the motion 

w 2 AI 

is to be transmitted by 
the direct contact of two 
profiles yy and Xx while 
rolling still continues be- 
tween aa and bb, the 
common normal to the 
profiles must continually 
pass through I, which is 
called the Pitch Point. 
Evidently as motion con- 
tinues, P will trace up 
a curve in space, whose 
form depends on the 
forms of the pitch curves 
and profiles. This curve 
is called the Curve or Locus of Action. The pitch curve, the 
profile, and the curve of action, are connected by the general 
law, that the line connecting the point of intersection P of the 
last two, with the point of intersection of the first and the line of 
centres, must always be normal to the profile at P. (See Fig. 46.) 




Fig. 46 



AXES OF ROTATION PARALLEL 



71 



Evidently then the profile cannot approach nearer to, nor recede 
further from, its centre of rotation than does the curve of action. 

The pitch curve is determined once for all when the nature of 
the angular velocity ratio is chosen. The curve of action is a curve 
fixed in space. If it be chosen of any desired form, the profile 
will then be determined. The most general proposition is as 
follows : let B (Fig. 46) be a centre of rotation, and BA a line 
fixed in direction from which angular position can be measured. 
Let bb be any given pitch curve, whose equation is known, cutting 
BA at I. Assume any curve of action cc, which will be fixed in 
space. We are then to find what curve xx fixed in the plane of bb, 
and therefore carried around B thereby, will have as a normal the 
line connecting its point of intersection with cc to /. The general 
solution of this problem is one of considerable complexity, and 
even in the simpler cases, the correctness of action of an assumed 
pair of profiles can be shown by methods far easier than its direct 
solution, but as the underlying law 
it is well to be kept in view. 

Definitions and Standard Dimen- 
sions. — Before proceeding to the 
deduction of any system of special 
profile forms, it will be best to 
give a few general definitions and 
dimensions, so that what follows 
may the more readily be under- 
stood. The teeth of gears are 
usually equally spaced along the / 
pitch curve, that is to say, the a ' 
distance measured along the pitch 
curve between its successive inter- 
sections with profiles facing in a 
given direction, will all be the 
same. One of these distances 
is called the Circular Pitch (Fig. 

47), and is denoted by F. It is measured usually in inches. 
The circular pitch must of course be the same on both of a pair 




Fig. 47 



72 MACHINERY OF TRANSMISSION 

of mating wheels. A fraction of a tooth being an impossibility, it 

must be an aliquot part of the perimeter of the pitch curve. 

C 
Hence if C be this perimeter, and iVthe number of teeth, P'= — • 

In good cut gears the circular pitch is equally divided between the 
tooth and the space, so that the thickness of the tooth on the pitch 
curve will be t= \P '. In rough cast gears the tooth is made a little 
smaller than the space, and the difference is called the Backlash. 

Another most important pitch dimension, which really should be 
used only in connection with circular gears, is the Diametral Pitch, 
denoted by P. It measures the number of teeth per inch of 
diameter of such a gear. Hence if D be the pitch diameter, 

P = — . But in a circular gear, C = irD, thus P' == -^r, and we 

obtain the relations, P= — ,, and P' = — . Since the size of tooth 

represented by a given diametral pitch is of fixed magnitude, we 
may apply the term to non-circular pitch lines, remembering that in 
these cases it means that if there were N teeth of such a size placed 
on a circular gear, there would be P teeth per inch of its diameter. 
That portion of the tooth lying without the pitch curve is called 
the Addendum, and is denoted by a. The portion lying within 
the pitch curve is a little greater than the Addendum, and the 
difference is called the Clearance. The angle through which a 
wheel turns while a tooth is in contact with its fellow is called 
the Angle of Action. That portion of this angle turned off while 
the point of contact is approaching the line of centres is called the 
Angle of Approach and the rest, the Angle of Recess. Those 
portions of the pitch curve intercepted by these angles, are called 
the Arcs of Action, Approach, and Recess. The Arc of Action 
must evidently be greater than the circular pitch. 

B. Velocity Ratio Constant, Circular Wheels 

In this case the pitch curves must evidently be circles, and / will 
be a fixed point on the line of centres. A curve of action being 
assumed, all the teeth will be alike. Upon the nature of the curve 
of action only will depend the various forms of teeth used. 






AXES OF ROTATION PARALLEL 



73 



(a) The Cycloidal System 

In this system the curve of action is a circle tangent to the pitch 
circle (Fig. 48). The solution of the general proposition in 
this case is of rather a 
complex nature (see 
Appendix III), but the 
special case where the 
radius of the pitch cir- 
cle is infinite, that is, 
where it becomes a 
straight line as in the 
ordinary rack, is simple 
enough to be inserted 
here. 

Let x and y (Fig. 49) 
be the coordinates of 
the point of contact P. 
Let x — f (y) be the 
equation of the profile, 
and x = (f> (y), be that 
of the curve of action, 
which in this case is, 
x = V2 ay — y-, " a " 

being the radius of the circular line of action. The tangent 
of the angle which the tangent to the profile at P makes with the 

dx 
axis of Y is — — • The tangent of the angle between OP and the 




axis of X is 



y 



dy 



But these are equal, hence, 

y _ y _ dx 



—f: 



X 

ydy 



V 



2 ay — y 



dy 



-1 y / — 

- = a vers 1 V 2 ay 

,2 a 



f+C, 



V2 ay — y 

which is the equation of a cycloid, with origin on the line of cusps, 
and at a distance C from a cusp. C is the variable parameter 
which determines the position of a tooth. 



74 



MACHINERY OF TRANSMISSION 



A simple method of finding the nature of the profile for the 
general case is to imagine a reversal of the relative motion. Bring 
the pitch curve and profile to rest by giving the whole system an 
equal and opposite rotation about B (Fig. 48) thus rotating the line 
SB, which carries the circle £ with it. If, however, the circle S 
be rotated about its own centre, its intersection with the profile 
will not be altered, hence we may allow it to roll upon the pitch 
circle, while S is carried around B. The path of any point on the 
circumference of circle S, is an epicycloid on circle B, and, / being 




Fig. 49 



the instantaneous centre, ZP must be normal to the curve described. 
If the circle *S lies within the circle B the curve described will be 
a hypocycloid, hence these curves may be used as profiles. The 
fact that any point in the plane of any curve rolled on the pitch 
curve will sweep up a correct profile, has led to the use of this 
feature as a basis for a tooth theory. It does not apply, however, 
in a practical way to all forms of tooth outlines in use. 

The relative positions of two profiles on a pair of pitch circles 
are shown in Fig. 50. The profile of circle B will be an epicycloid 
ee since the circular curve of action lies without its circumference. 



AXES OF ROTATION PARALLEL 



75 



The profile of A will evidently be a hypocycloid kk . The point of 
contact is P, and the common normal PI continually passes through 
/, a fixed point. The circular curve of action is called the De- 
scribing Circle. Action will take place, i.e. driving at the required 
constant angular velocity ratio will take place, on one side of the 
line of centres only. If A be the driver while rotating counter- 
clockwise, this will be on the left side as shown. After P has 




Fig. 50 



passed the point / where the cusps come together, contact will no 
longer be between e^e and /i k, but between the second branches 
e e' and A k' beyond the cusps, and furthermore, the tendency 
would be for hji 1 to separate from e e' instead of driving it for- 
ward by pushing. 

By using a describing circle within each of the pitch circles, the 
angle of action may be extended to both sides of the line of 
centres. It will then be divided into an angle of approach plus an 



7 6 



MACHINERY OF TRANSMISSION 



angle of recess on each wheel. Fig. 5 1 is a diagrammatic represen- 
tation of the manner of forming the tooth outlines. Point P of 
describing circle S traces the face PC of the tooth of B, and the 
flank PD of the tooth of A ; while the point Q of the circle T 
traces the face QE = face DX of the tooth of A, and the flank 
QF= flank CFof the tooth of B. We see that in this case the locus 
of contact of two teeth will be PMINQ. The arcs of action CIF 
or DIE will be composed of arcs of approach DI or CI, plus arcs 




Fig. 51 



of recess IE or IF. The arc of approach depends upon the 
length of the follower's face only, and that of recess upon the 
length of the driver's face. 

Inside or Annular Wheels. — When one pitch circle lies within 
the other, the teeth of the larger wheel must be cut on the inside 
of an annulus or ring. In this case the direction of rotation of 
the wheels will be the same. The tooth outlines are generated in 
precisely the same manner as in outside gearing. The flanks of 



AXES OF ROTATION PARALLEL 



77 



the teeth will be epicycloids, and the faces hypocycloids ; in 
fact, the spaces correspond exactly with the teeth of outside 
gearing of the same size. Inside gearing will in general run more 
smoothly than outside gearing, but owing to the difficulty of con- 
struction they are seldom used. When rotations in the same 
direction are desired, an idle wheel is introduced. 

Racks. — A rack is simply a portion of a wheel of infinite diam- 
eter, hence the pitch curve of a rack is a straight line tangent to the 




Fig. 52 

pitch circle of its mating wheel. Both the faces and flanks of the 
rack teeth will be cycloids, and if the two describing circles are 
equal, the faces and flanks will be alike. 

Interchangeable Wheels. — If we wish to make a set of wheels 
any one of which will gear with any other, we must use the same 
size of a describing circle for all the faces and all the flanks. The 
size of the describing circle depends on the properties of the 
hypocycloid. If the diameter of the describing circle is less than 



7% 



MACHINERY OF TRANSMISSION 



one-half that of its pitch circle (Fig. 52), the flanks of any tooth 
will be less converging than the radii of the pitch circle ; if equal 
to half the diameter (Fig. 53), the flanks will be radial; and if 
more than half the diameter (Fig. 54), the flanks will be more 
converging than the radii of the pitch circle. Since converging or 
even radial flanks weaken the tooth at its root, which is the place 
where the greatest strength is needed, we should not have the 
common describing circle of a set of wheels much greater than 







Fig. 53 

half the diameter of the smallest wheel of the set. The Brown 
and Sharpe Mfg. Co. uses a describing circle which is half the 
diameter of a gear of 15 teeth, and cuts down to 12 teeth. Since 
the diameter of a wheel of 15 teeth is 

P P 

the diameter of the describing circle will be 

'-* 



AXES OF ROTATION PARALLEL 

Formulae for proportioning Teeth (Fig. 55). 
Let N = the number of teeth, 
a = the addendum, 

/ = the thickness of the tooth on the pitch curve, 
/ = the clearance at the bottom of a space, 



79 




FIG. 54 



d = the working depth of a tooth, 
d -{-/ = the whole depth of a tooth, 
D — the pitch diameter, 
Z>' = the outside diameter, 
P' = the circular pitch, 
and P= the diametral pitch. 



So 



MACHINERY OF TRANSMISSION 



Then P=^ y 



P' = 



P* 



a =± = £ = .31*3 P, 

d = 2a, 

N 



P = 



D' 



N 



£>'=D+2a = — + ^- = 

pTp 

t— i/ 3 ' — — 




Fig. 55 



/= 



l /_ l pi — 



2oP' 



P ' 20P 



= l |1 



= 2.i57i 



Note. — /*' and / are measured on the arc of the pitch circle, and not on 
the chord. 

In cutting gears with the rotary cutter, the important formulae 

are (i), P = — • This gives the relation between size, pitch, and 

number of teeth. (2), D' = — ^t— , which gives the size to which 

the blank must be turned, and (3), d+f= * " -, which is the 

depth the cutter must be run into the blank. These are all that 
are required in simple spur gear cutting. In cut gears the tooth 
is made equal to the space, and hence the backlash is zero. 

Approximate Methods of drawing Teeth. — The most exact method 
is by using the average radius of curvature of the cycloidal arc and 
approximating it by means of the arc of a circle. George B. Grant * 
has constructed a table giving radii and positions of centres, the 



* " Odontics, or the Theory and Practice of the Teeth of Gears." Lexington 
Gear Works, Lexington, Mass. 



AXES OF ROTATION PARALLEL 



81 



radius of the describing circle being one-half that of a gear of 12 
teeth. This he calls the " Three-point Odontograph." To use 
the table draw the pitch circles, and divide them up into the tooth 
intervals. Draw the circle of face centres at a tabular distance 
(dis.) inside the pitch circle, and the circle of flank centres at a 
tabular distance outside the pitch circle. Draw faces and flanks 
with tabular radius (rad.). Note must be taken of the algebraic 
sign of the radius. For 12 teeth the flank radius is infinite, and the 
flanks are straight. For less than 1 2 teeth the flanks are convex, and 




for more than 12 they are concave. Fig. 56 shows the complete 
construction for the case of 15 teeth, 3-pitch. No odontographic 
table,, such as is shown above, has been worked out for a system 
where the describing circle is half the diameter of a gear of 15 
teeth, and is used by the Brown and Sharpe Mfg. Co., but suffi- 
ciently accurate methods for finding the approximate radius and 
position of the centre by graphical construction are given in " A 
Practical Treatise on Gearing," published by the company. 
Draughted teeth are used principally by the pattern-maker. 



82 



MACHINERY OF TRANSMISSION 



THREE-POINT ODONTOGRAPH 

Standard Cycloidal Teeth 
From a Pinion of Ten Teeth to a Rack 







For One Diametral Pitch. 


For One-inch Circular 


Pitch. 






For any Other Pitch, divide 


For any Other Pitch, 


multi- 


Numb 
Te 


ER OF 
2TH 


by that Pitch 


. 


ply by that Pitch. 






Faces 


Flanks 


Faces 


Flanks 


Exact 


Interval 


Rad. 


Dis. 


Rad. 


Dis. 


Rad. 


Dis. 


Rad. 


Dis. 


IO 


IO 


1.99 


.02 


—8.00 


4.OO 


.62 


.OI 


—2-55 


I.27 


II 


II 


2.00 


.04 


—II.05 


6.50 


.63 


.OI 


—3-34 


2.07 


12 


12 


2.0I 


.06 


00 


00 


.64 


.02 


00 


00 


13-5 


I3-H 


2.04 


•07 


15.IO 


9-43 


.65 


.02 


4.80 


3.OO 


!5-5 


I5-I6 


2.IO 


.09 


7.86 


346 


.67 


•03 


2.50 


I.IO 


17-5 


I7-I8 


2.14 


.11 


6.13 


2.20 


.68 


.04 


i-95 


•70 


20 


19-21 


2.20 


•*3 


5.12 


i-57 


.70 


.04 


1.63 


•50 


23 


22-24 


2.26 


•15 


4-5° 


i-i3 


.72 


•°5 


1-43 


.36 


27 


25-29 


2-33 


.16 


4.10 


.96 


•74 


•05 


1.30 


.29 


33 


30-36 


2.40 


.19 


3.80 


.72 


.76 


.06 


1.20 


•23 


42 


37-48 


2.48 


.22 


3-5 2 


.63 


•79 


•°7 


1. 12 


.20 


58 


49-72 


2.6o 


•25 


3-33 


•54 


.83 


.08 


1.06 


•17 


97 


73-M4 


2.83 


.28 


3-i4 


•44 


.90 


.09 


1. 00 


.14 


290 


I45-3 00 


2.92 


•3i 


3.00 


•38 


•93 


.IO 


•95 


.12 


00 


Rack 


2.96 


•34 


2.96 


•34 


•94 


.11 


•94 


.11 



Radius of fillet at the root of tooth may be taken as one-sixth the distance 
between tips of teeth. 

General Expressions for Angles of Action. — Let A and B 

(Fig. 57) be the centres of two pitch circles whose radii are R\ 
and R 2 . Let r x and r 2 be the radii of the corresponding de- 
scribing circles. Draw the addendum circles KL and GH, 
where a x and a 2 are the addenda. Suppose the wheels revolve 
so that the teeth move from right to left, with A as driver. We 
are to find the value of a, the angle of approach of A. The point 
P, at the intersection of the describing circle and the addendum 
circle, is the first point of contact. Draw PS and PB. Then in 
the triangle PSB, all the sides are known, for PS=r 1} SB=R 2 -\-r lf 



AXES OF ROTATION PARALLEL 



83 



and PB = R 2 + #2- I n an y sucn triangle we can find the angle 
from 



tan 






-(P 2 + r 1 )l\x-r 1 l 



(^2 + ^2)! 
where ^ is half the sum of the three sides, or 

x = P 2 + r 1 + 



2 


.4 






^v 


?! / 
B 



H 



Hence, 



6=2 tan" 1 



0=2 tan" 



1 


?( 


2 ^ 2 + <^2 N 


) 






/ (2 R* 


+ 2 ; 
2 


"l + ^>V 2 


*i- 


« 2 N 




'A 


A 


2 


/ 


/ 


! 


^2(2 


•ff 2 + <2 2 ) 









«a) 



84 MACHINERY OF TRANSMISSION 

Since arc/£> = arc ZP = arc 7C, a = 0^-, y = f ±. 

#1 #2 



Hence, a = ^ tan" 1 J-_ — 

^1 * ( 2 /to 



« 2 (2^ 2 + a 2 ) 



i?! ^(2^+2 rj + a 2 ) (2 r x — a 2 ) 

This is the most general expression for the angle of approach 
of the driver. For standard teeth the following arbitrary propor- 
tions can be substituted : 



a x = a 2 = — • 



15 



Substituting these values, we get 



3 tan" 1 



2PR X 

And since 2 PR = N, the number of teeth, 



V(-*+3 + ?X3-?) 



« = ^tan- 1 \/^r 



4W + 1) 



2^+17) 

In exactly the same way we get the angle of recess of the 
follower : 

8 Ij tan - W ' 4W+x) 
JVi \ 13(2^+17) 

Since y = <*— , the angle of approach of the follower is 
^2 

? = ^ 2 tan V I3(a ^ +I7 )- 

And finally, 



fl I5 tan-J 4W+0 



)_ 
7)' 



In all cases the arcs of action, which are .£,(«+/?) =iP s (y+8), 
must be greater than the circular pitch. When the standard 
addendum and describing circle are used, the least number of 



AXES OF ROTATION PARALLEL 85 

teeth on a wheel is not decided by the above conditions, but by 
the undercutting which occurs when the pitch circle becomes less 
than twice the describing circle. When, however, the addenda 
and describing circles are functions of their respective pitch cir- 
cles, then definite limiting cases can be worked out. For exam- 
ple, let r x = KR X , r 2 = CR 2 , a x =pR x , a 2 = qR 2 , where K, C, p, 
and q are known constants. Then the angle of C is 



a + fl=,X*r*4 



(2^+2 KR X + qR 2 ) (2 KR X - qR 2 ) 



+2 c^u a -4 >*'<»+» 



Ri ^(2^+2 CR 2 +pR x ) ( 2 CR 2 -pR x ) 

The limiting case will be when a-\-fi equals the pitch angle, 



2 7T 



or - — , and in general it cannot be less. Hence, writing the 

■N . N 

inequality, and remembering that R x — — ^, etc., 



^tan-^ 



:JV. 2 (2 + q) + 2 KN X \ \ 2 KN X - qN, 



VPNH2 +fi) 
\N,{2 +p) + 2 CJV,J S 2 CN 2 -/Ai| > *• 

Having given all quantities except N 2 , this latter can be found 
by a process of trial and error as that whole number giving the 
nearest result larger than ir. The still further limitation must then 
be investigated as to whether the epicycloids forming the two 
faces of a tooth do or do not cross before reaching the addendum 
circle. The whole problem of limiting numbers is one which can 
best be studied by graphical trial processes. 



(b) The Involute System 

The simplest and in many respects the best curve for the out- 
line of a tooth is the involute of a circle (Fig. 58). In this system 
the curve of action is a straight line passing through the pitch 



86 



MACHINERY OF TRANSMISSION 



point. Here again the profile for the special case where the pitch 
curve is a straight line can be deduced very simply from the gen- 
eral law. Here x =/ (y) is the equation of the profile, and that 



of the curve of action is x 
gent of (Fig. 59). Hence, 

y _ y _ dx 
x y dy 

m 
y = — mx + C, 



^(y)~m> m b em S equal to the tan- 



dy 



dx 




Fig. 58 

which is the equation of a straight line at right angles to the curve 
of action, C being the arbitrary parameter giving the position of 
the tooth.* 

Again resorting to simpler methods of procedure, imagine a 
reversal of relative motion, thus bringing the pitch curve and pro- 
file to rest, and rotating the curve of action about the axis of the 



* For the solution of the general case, see Appendix IV. 



AXES OF ROTATION PARALLEL 



87 



wheel. Then, as any motion of the curve of action in the direc- 
tion of its own length will not affect its intersection with the pro- 
file, we may consider it as rolling upon a circle drawn concentric 
with the pitch cir- 
cle and tangent to 
the curve of ac- 
tion. (See Fig. 
58.) The part 
of any point of 
the line will be 
an involute of 
this inner circle. 
The curve of ac- 
tion itself is a 
normal to the 
involute, and as 
it always passes 
through the point 
/, the tooth law 
is satisfied. The 

relative positions of two profiles are shown in Fig. 60. In this 
let A and B be the centres of two pitch circles tangent at /. 
Through I draw a line WW making an angle with the common 
tangent to the circles. From A and B drop perpendiculars AE 
and BF upon WW, and with these as radii draw circles concen- 
tric with the pitch circles, and therefore tangent to the line of 
action. Now suppose the line WW to be pushed in the direction 
of its length so as to drive circle CEO by friction at E, and 
circle DFQ by friction at F. Then will the pitch circles roll 

IA 
upon one another also, for vel. /= vel. E when considered a 

IB EA 

point of A, and vel. / = vel. F — when considered a point of B. 

I A _IB , FB 

But -jrz — ~J^E>, and vel. E =vel. F. Hence vel. /when considered 

a point of A is equal to vel. / when considered a point of B, or the 
pitch circles roll at /. Now if we consider any point P of the line 




Fig. 59 



88 



MACHINERY OF TRANSMISSION 



WW, it will sweep up an involute GPH of the circle OEC in the 
plane of A, and an involute MPN oi the circle DFQ in the plane 
of B. The curves will always be tangent at the common generat- 
ing point P, and their common normal will be the line WW, which 
always passes through a fixed point / on the line of centres. The 
circles OE C and QFD are called Base Circles. 




Fig. 60 



In involute teeth the addenda a cannot be of greater lengths than 
those which cause the last points of contact to occur at E and F. 
In other words, the radii of the addendum circles cannot be greater 
than BE and AF. In this case the action ends at the tips and at 
the roots {i.e. at the point where the involute comes down to the base 
circle) of the teeth. If the tooth of B, for instance, were made 
any longer, action would continue beyond the root of A, that is 
along the second branch G X H X of the involute beyond the cusp as 
shown at P x in Fig. 60. Hence we see that the maximum line of 



AXES OF ROTATION PARALLEL 89 

action is EF. If the tooth were made longer than the theory would 
indicate, not only would correct angular velocity transmission end 
at E, but the prolonged tooth would actually interfere with the 
other. This arises from the fact that the curvatures of the por- 
tions mathematically in working contact are in the same direction, 
and the radius of curvature of the tooth of B being FF 1} while 
that of A is only EF 1} the latter would lie within the former, as 
would also the cusp of A at G±. 

A valuable feature of the involute tooth outline is that which 
allows the shafts of the two wheels to be separated slightly without 
affecting the constancy of the angular velocity ratio. In this case 
the base circles remain the same as before, and hence their invo- 
lutes are unchanged. What is changed by the separation of the 
axes is the pitch circle. Since the ratio of the radii of these 
always equals the ratio of the radii of the base circles, the former 
ratio will remain unchanged. It is readily seen that the angle 0, 
or the " obliquity," is changed also. Since the new pitch circles 
do not cut the teeth in the same positions as the old, the tooth 
will not equal the space on any pitch circle except the one origi- 
nally designed, but there will be back lash. Care must be taken 
that the shafts are not so far separated as to make the angle of 
action less than the pitch angle. 

Standard Involute Tooth. — The draughted tooth is usually one 
having an angle of obliquity of 15 . Other proportions are the 
same as for cycloidal teeth. The Brown and Sharpe Mfg. Co. 
makes its cutters with an obliquity of 14J . If the obliquity is 

1 5 , and the standard addendum — is used, it will be found too 

long to work without interference on anything under 20 teeth on 
equal wheels. Therefore the involute is used as far as possible, 
and the remainder of the tooth outline is made an epicycloid to 
work on a radial flank within the base circle. This is known as 
the correction for interference. 

The Invplute Odontograph. — In the involute odontograph of 
George B. Grant the centres are taken on the base circles. These 
latter may be drawn tangent to a line of action with an obliquity 



9 o 



MACHINERY OF TRANSMISSION 



THREE-POINT ODONTOGRAPH 

Standard Involute Teeth 

Obliquity 15 





For One Diametral Pitch. For 


For One-inch Circular Pitch. 


Number of 
Teeth 


any Other Pitch, divide by that 
Pitch. 


For any Other Pitch, multiply 
by that Pitch. 




Face Rad. 


Flank Rad. 


Face Rad. 


Flank Rad. 


IO 


2.28 


.69 


•73 


.22 


II 


2.40 


•«3 


.76 


.27 


12 


2.51 


.96 


.80 


•31 


13 


2.62 


I.09 


^3 


•34 


H 


2.72 


1.22 


.87 


•39 


J 5 


2.82 


i-34 


.90 


•43 


16 


2.92 


1.46 


•93 


•47 


17 


3.02 


1.58 


.96 


.50 


18 


3.12 


1.69 


•99 


•54 


19 


3.22 


1.79 


1.03 


•57 


20 


3-32 


1.89 


1.06 


.60 


21 


3-41 


1.98 


1.09 


•63 


22 


3-49 


2.06 


1. 11 


.66 


23 


3-57 


2.15 


113 


.69 


24 


3-64 


2.24 


1. 16 


.71 


25 


3-7i 


2-33 


1. 18 


•74 


26 


378 


2.42 


1.20 


•77 


27 


3-85 


2.50 


1.23 


.80 


28 


3-92 


2-59 


1.25 


.82 


29 


3-99 


2.67 


1.27 


•85 


30 


4.06 


2.76 


1.29 


.88 


31 


4.13 


2.85 


i-3* 


.91 


32 


4.20 


2-93 


i.34 


•93 


33 


4.27 


3.01 


1.36 


.96 


34 


4-33 


3-09 


1.38 


•99 


35 


4-39 


3.16 


i-39 


I.OI 


36 


4-45 


3-23 


1.41 


1.03 


37-4o 


4.20 


1-34 


41-45 


4.63 


1.48 


46-51 


5.06 


1.61 


52-60 


5-74 


1.83 


61-70 


6.52 


2.07 


71-90 


7.72 


2.46 


91-120 


9.78 


3-ii 


121-180 


I3-38 


4.26 


181-360 


21.62 


6.88 



In all cases the centres are on the base circles, 
method described in text. 



Draw rack by special 



AXES OF ROTATION PARALLEL 



91 



of 1 5 , or they may be drawn within the pitch circles at distances 
equal to .017 — . It is necessary to round off the points of the 

teeth in some cases to prevent interference. This makes it 
impossible to compute the positions of the centres, hence careful 
draughting is resorted to. The involute is plotted, corrected for 
interference, and the circle most nearly coinciding with it is 
found by trial. This is done on a very large scale so as to reduce 




Fig. 61 



the errors to a minimum. Separate curves are required for face 
and flank up to 36 teeth, but above that one curve is sufficient. 
The sides of the rack teeth are straight lines drawn at an angle 
of 1 5 with the perpendicular to the pitch line, but the tips must 
be corrected for interference by rounding them off. Hence draw 
the outer half of the rack face from a centre on the pitch line 
with a radius equal to 2.1 divided by the diametral pitch or .67 
multiplied by the circular pitch. An example is shown in Fig. 61. 



9 2 



MACHINERY OF TRANSMISSION 



Angles of Action. — Let A and B (Fig. 62) be the centres of two 
pitch circles, whose radii are B 2 an d R Y . The radii of the corre- 
sponding base circles are r 2 and r 1} and the addenda are a 2 and a L . 
The obliquity is d. Action will begin at P, where the addendum 




Fig. 62 



circle of B cuts the line of action. The angle of approach of B 
is JBC=y, and it can be calculated as follows : 



y + s^^J^ + atf-r*, 

s =V(B 1 + a 1 y-^-y 



Or 



s = V {Bi + a x y - B{ cos 2 - Rx sin 6. 



AXES OF ROTATION PARALLEL 93 

But since IP == s = GH, 






„ V(i?i + a,) 2 - i 1 ?! 2 cos 2 6 — R x sin (9 

Hence, y = - — v * l • 

' Y R x cos e 

In the same way the angle of recess of A is 



q = V(^ 2 + atf - Rj cos 2 9 - R 2 sin 6 
P R 2 cos 9 

r> 

Since a = y ~, the angle of recess of B is 

Ri 



a = v (^1 + ^i) 2 - Rf cos 2 fl - R x sin 
RoCosO 



and that of approach of A is 



g = V(^2 + a 2 ) 2 - i? 2 2 cos 2 9 - #, sin (9 

^cos e 

If we use the standard values for a and 9, i.e. a = — and 

P 
9 = 15°, then 

^ x + jjY- .933 *i 2 - .2588 ^ 

7= .966 ^ 

The above does not consider the effect of interference. The 
angles of approach and recess as limited thereby can be simply 
deduced as follows. In this case s = R 2 sin 9, and 

s R 2 sin 9 R 2 . n 

y = ~ = -± = -l tan 9. 

1 j\ R 1 cosd R x 

Similarly, = ^ tan 9. 

R2 

And as a =y —, 

1 R2 

a = tan 0, 8 = tan 9. 



94 



MACHINERY OF TRANSMISSION 



(c) The Pin-tooth System 

The pin-tooth system, though totally different from either of 
those previously described, can be deduced from the cycloidal 
system. Suppose in this case one describing circle becomes equal 
in diameter to the pitch circle within which it lies, while the other 
is omitted. Since there is but one describing circle, the teeth will 
consist of merely epicycloidal faces on one wheel, and hypocy- 




cloidal flanks on the other. But the hypocycloidal portions will 
vanish, these reducing in fact to a point which is the describing 
point itself. This point will then work correctly with the epicy- 
cloid of the other wheel. The combination of P' with P'Q'S' 
(Fig. 63) is a true cycloidal pair, the line IP being the normal 
to both. A mathematical line at P' would drive the epicycloid as 
far as 7, but beyond / there would be no positive driving. If now 



AXES OF ROTATION PARALLEL 95 

in place of P' we put a cylindrical pin of radius r 0t the line IP 1 
will always be normal to the pin. Finally if we replace the epicy- 
cloid by a curve PQ at a constant normal distance from it, the 
line IP will be normal to this also. Hence the combination of the 
cylindrical pin and the parallel to the epicycloid will work as a 
pair of gear teeth, the common normal always passing through /. 
The proof of the correct action of teeth in this system can also 
be studied by imagining the outlines as being formed by the "con- 
jugate method." It is evident that if any form of tooth outline 
be assumed for the pinion, a corresponding correct tooth outline 



Fig. 64 

can be devised for the wheel, if only the mathematical conditions 
are to be fulfilled. For if the required angular velocity ratio take 
place, this outline will be the envelope of all positions of the 
assumed outline when referred to the wheel. Assuming one tooth 
as a circle, the other must be the parallel to the curve traced out 
by that circle's centre (Fig. 63). 

The point of contact of the pin and tooth lies on the line P'l 
at the point P. The curve of action will be the curve IPF. This 
curve is evidently the locus of a point of a line which moves with 
one extremity on a circle, and always passes through a point on 
the circumference of the circle. P' is the moving point, P the 



9 6 



MACHINERY OF TRANSMISSION 



tracing point, and / the fixed point. The distance PP — PP' = 
constant = r is the radius of the pin. Such a curve is known as a 
limacon (see Fig. 64), and was first investigated by Pascal. If 
we take / as the pole, and the diameter through / as the reference 
line, the equation of the limacon will be 

r — 2 a cos 6 — b. 

It is evident that action will take place wholly upon one side 
of the line of centres, and hence will be wholly approaching or 

receding, depend- 
ing upon whether 
the pin wheel is 
driving or following. 
Since the friction of 
approach is greater 
than that of recess, 
it is best to give the 
pins to the follower. 
In the case of 
inside or annular 
gears, the tooth 
curves will be par- 
allels to the hypo- 
cycloid. One case 
is of particular 
interest ; namely, 
when the pin wheel 
is half the diameter 
of the gear. The tooth outlines will then be parallel to the 
straight line hypocycloid, or will themselves be straight (Fig. 65). 
If the pins are set on the circumference of the pitch circle itself, 
the action will be defective while the line of centres is being passed. 
This is best seen by considering the epicycloid as being generated 
by a point of a line which rolls upon its evolute. Let Oabcd, etc., 
be the epicycloid, and OX its evolute (Fig. 66). Then the parallel 
to the epicycloid is generated by another point of the same rolling 




Fig. 65 



AXES OF ROTATION PARALLEL 



97 



line. But this latter will come down to the evolute at b', while the 
point tracing the epicycloid is still at b, and by the time the latter 
reaches O the parallel curve will have risen again to O', forming a 
cusp at b'. So the curve O'a'b'c'd'e'f is the parallel to the epicy- 
cloid Oabcdef. But the branch O'b' would work on internal contact 
with the pin when near the cusp, and hence is impracticable. The 
only way to prevent this failure of action is to use some other curve 
than the epicycloid to parallel. If the pins are set a suffi- 




Fig. 66 

cient distance within the pitch circle, the resulting curve, the epi- 
trochoid, can be paralleled without a cusp. (See Fig. 67.) 

Approximate Formula for Angles of Action. — In Fig. 68 the 
first point of contact will be where the addendum circle cuts the 
limacon IF. The circumference of the pin, the tip of the tooth, 
and the line (?/also pass through this point. Construct the tooth 
curve PC. Then angle IBC = a is the angle required. Let r 
equal the radius of the pitch circle of A, R that of B, r the radius 



9 8 



MACHINERY OF TRANSMISSION 



of the pin, and a the addendum. The irregular action at / is dis- 
regarded, the mathematical action only being considered. 
s = BP=R + a, 
P = angle QIA, 
e = angle QIB, 
P + t= 180 , 
cos -J- e = cos \ (i 8o° — P) = sin i /?. 




Fig. 67 



Call //> = > Then fQ=y + r , also PB = s 9 BI=R, 

IA= QA = r. Then 

, x(x — s) 

where # = |(^? + y + j) ; 



hence, cos 2 J c = 



(P+y + s)(P+y-s) 
yR 



= sin 2 ip. 



AXES OF ROTATION PARALLEL 
But we also have 



99 



where 
hence, 



sin i 



J(x>-r)(x'-y-r ) 
P-M r(r +y) 

x' = i(r+r+r +y) ; 



&t ip = *"-*- y) - 




Fig. 68 



Equating values of sin 2 \ /?, we get 

(R+y + s)(R+y-s) 2r-r -y 



Reduction gives 



yR 



Rr, \/-R* RW 

> + r)^\ r + R ^ 4 (r + Ry 



2(R 

which is in terms of known quantities. 

LofC. 



100 



MACHINERY OF TRANSMISSION 



Now sin J 



Hence, 



Uy + n) _,_--i fK-y+^o) 



or = 2 sin -1 { 

Arc QI=rO = Arc /£>, 

Arc /C = Arc /Z? — r (nearly) . 

Arc IC= rO — r (nearly) . 

.^(^l-r, 
I r j 



2 r sin" 



-Rn 



2(R + r) 



+ 



v 



.y 2 -^ 2 , R 2 r 2 
r — -f- 



r + R 4{R + r)' 



+ r 



— r 



(d) Special Forms of the Above, Twisted Gears 

If a pair of spur gears be divided by a series of parallel planes 
perpendicular to the axes of rotation into a set of thinner gears, and 
if each pair of these on both wheels be given an angular displacement 

with respect to the pair 
preceding, we will have 
what is called a pair of 
stepped gears (Fig. 69). 
They will, of course, 
work together exactly as 
did the original wheels, 
but with the advantage 
that the number of teeth 
passing the line of cen- 
tres during a given an- 
gular displacement will 
be increased as many 
times as there are lam- 
inae. In this way the 
effective number of 
Fig. 69 teeth may be increased 







II 1 1 1 


1 1 1 II 


II 1 1 


II II II 


1 II 1 1 


II III 


II 1 II 


1 1 II 



AXES OF ROTATION PARALLEL 



IOI 



without reducing the size of a tooth. Since, when the point of 
contact between two teeth is passing the line of centres, driving 
is secured with pure rolling, the pair of teeth is at that instant 
acting at its best. Hence by subdividing a gear in this way the 
action will be smoother, while the strength will not be impaired. 

If the number of laminae is made infinite, the effect will be the 
same as if we gave the gear a uniform twist throughout its length 




Fig. 70 



and about its axis of rotation. These are called Twisted Gears, 
and are much used on account of their smooth action. The tooth 
surface can also be imagined as swept up in a manner identical 
with those already mentioned. We have so far considered the 
whole subject of spur gearing as a problem in plane geometry, but 
when successive sections of our gear differ either in form or in 
position, we must employ the geometry of three dimensions. 



102 



MACHINERY OF TRANSMISSION 



According to this last the cycloidal tooth would be a surface 
generated by an element PR (Fig. 70) of a right circular cylinder, 
rolling within and without a pitch cylinder, and the involute tooth 
a surface swept up by a line PR of a plane (Fig. 71), rolling 
between base cylinders, the generating line being parallel to the 
line of contact TS between the plane and the base cylinders. 




Fig. 71 

Now in the twisted gear the tooth surface is swept up by a helix 
PQ (Fig. 70) of uniform pitch on the describing cylinder, where 
cycloidal teeth are considered, or by a line PQ (Fig. 71) of a 
plane, the generating line being oblique to the plane's lines of con- 
tact with the base cylinders, in the involute system. In either case 
it will be seen that the elements will no longer be straight lines 
but helices, and the tooth surfaces more or less complicated 
helicoids. 



AXES OF ROTATION PARALLEL 



IO3 



In such a gear the sections of the tooth made by planes normal 
to the axes of rotation, such as CD (Fig. 72), will give true profiles 




according to any of the systems already mentioned ; but if cut on a 
milling machine with a rotary cutter, the form of this cutter should 
not be that of the true tooth, but of the normal section of the tooth 
EC, perpendicular to its helical elements. Such teeth are usually 




cut with the same cutters as are used in cutting spur gears, and the 
resulting tooth form cannot be exactly correct. However, the 



104 MACHINERY OF TRANSMISSION 

difference is so slight as to be inappreciable, and the action is in all 
cases extremely smooth, for there is always a point of contact on 
the line of centres, provided the twist displacement is at least 
equal to the circular pitch. 

In a plain twisted gear the normal pressure between two 
teeth can be resolved into a useful component tangent to the 
pitch surface and normal to the plane of the axes, and a useless 
component of end pressure. The components of end pressure can 
be neutralized by placing two twisted gears on the same axis with 
equal twists in opposite directions. (See Fig. 73.) 

(<?) Strength of Spur Teeth 

The actual load which is applied to a tooth at the instant of 
rupture is the same as that applied to one end of a beam which is 
fixed at the other. If w is the width, d the depth, and / the 
length of such a beam, 

where F is the breaking load and K a constant of the material. 
In the case of the tooth, w=f, the "face" of the gear, or the 
length of the tooth parallel to the axis, 

d= — = — , and /==2<z=— . 
2 2P P 

Hence, F = K — =^ = C — • 

' 4P 2 X 2 P 

The value of C is given by Grant as 11,000, but for actual running 
conditions a factor of safety of 10 is introduced, so that the work- 
ing load is 

^ = 1100^. 

This is for rough cast gears, where the whole load may be borne 
by one tooth. If two teeth are always in contact, it may be safe 
to allow twice the working load. This can only be done when the 
tooth outlines are correctly formed. 



AXES OF ROTATION PARALLEL 



105 



The horse-power which a gear will transmit can best be com- 
puted from the following empirical formula : 

H. P. = .I2^, 

where / is the " face " of the wheel, d the pitch diameter in 
inches, and n the number of revolutions per minute. This is for 
cast gears ; for the best cut gears we allow 

fVdn 



H. P. = .40 



p2 



C. Velocity Ratio Variable, Non-circular Wheel 



The most general practical case is that of two curves 
form so related that they roll upon one another while 
about fixed centres. If 
one curve is chosen of 
definite form, the mating 
curve can always be con- 
structed by the method 
of Fig. 74. Let FE be 
any curve rotating about 
B. Let A be the other 
centre. / is the intersec- 
tion of the given curve 
with the line of centres. 
Step off small arcs la, aa, 
aa, aa, etc., along FE. 
With centre B draw cir- 
cular arcs ac, ac, ac, etc., 
intersecting AB in c, c. 
With radius Ac draw cir- 
cular arcs cb, cb, cb, and 
lay off distances lb, bb, 
bb, equal to la, aa, aa. 
Then will Ibbbb be the 
required curve. If the 



of any 

turning 




io6 



MACHINERY OF TRANSMISSION 



curve FE is closed, the curve CD is not necessarily closed, but 
A can be chosen so that it will close. 




Fig. 75 



(a) Elliptic Gears 

The only pair of like closed curves which will roll upon one 
another while rotating about fixed centres, and make a complete and 

practical revolution, is a pair of 
equal ellipses. Let Fig. 75 rep- 
resent two such ellipses, with foci 
at A and C, and at B and D. 
Suppose them to have been origi- 
nally tangent at E and F (the 
extremities of the major axes), 
and to have rolled until tangent 
at /. Draw I A, IB, IC, and ID. 
Arc EI equals arc FI, and since 
the ellipses are exactly alike, 
CI=BI, and AI=DI Also 
A C = DB. Hence triangle AIC = triangle DIB, and angle 
AIC = angle DIB. But by the property of the ellipse, angle BIT 
= DIT 1 , where TIT X is the common tangent at /. Also by the 
equality of the ellipses, angle BIT= angle CIT. Hence, as the 
sum of the angles about / on each side of AIB or CID are 
equal, these latter must be straight lines. Furthermore by the 
property of the ellipse 
CI + IA= GE. But 
DI=AI. Hence DC 
= GE constant. So if 
we pivot the two ellipses 
at their two foci A and 
B, at a distance apart 
equal to their major axes, 

they will roll together without slipping. The free foci may, if 
required, be connected by a link, since CD = AB = GE. 

Let the ellipse whose axis is A be the driver, turning with a 




Fig. 76 



AXES OF ROTATION PARALLEL 



107 



constant angular velocity wj. When the ellipses are in the position 
shown in Fig. 76, we will have 

t _ w i _ BE — ( T + e ) a _ I + e 
o> 2 ^G (1 — e) a \ — e 

where a is the semi-axis major, or \ GE, and where e is the 
eccentricity. When the ellipses are as in Fig. 77, we have 



_ <°i _ BH _(* — e)a_i — <? 
w 3 AE (1 + e) a 1 -f e 



Hence, 



77 o> 2 (1 - ^) 2 




FIG. 77 

This quantity Z is generally the one assumed in the design of a 
pair of elliptic gears. It is the ratio between the maximum and 
minimum angular velocities of the follower when the speed of the 
driver is constant. From the last equation we can find the value 
of e from the given value of Z, for it gives 



e = 






In some cases we assume the ratio of minor to major axes, or -= K. 

Then e is calculated from its defining equation, viz., b 2 = a 2 (1 — ^), 
which gives 



-4 



= Vi-X 2 . 



io8 



MACHINERY OF TRANSMISSION 



In addition to Z or K we may assume the number of teeth, and 
the size of a tooth, that is the circular or diametral pitch. Let s 
represent the length of an elliptic quadrant. Then 



N 



P' = ^ 



-kN 



AP 



TABLE OF COMPLETE ELLIPTIC FUNCTIONS, E 



(••;) 



« 


E 


e 


E 


e 


E 


e 


E 


.OO I 


5708 


25 


1-5459 


•5° 


I.4674 


•75 


I.3I83 




OI I 


5707 


26 


1-5439 


•5" 


I.4630 


.76 


I.3I02 




02 I 


5706 


27 


1.5418 


•52 


I4585 


•77 


I.3020 




03 I 


5704 


28 


1-5395 


•53 


"4539 


.78 


1.2936 




04 I 


570I 


29 


J-537" 


-54 


"4493 


•79 


1.2852 




°5 " 


5698 


30 


"•5347 


•55 


"•4445 


.80 


1.2762 




06 1 


5 6 94 


31 


1.5322 


.56 


"•4395 


.81 


1/2671 




07 1 


5689 


32 


1.5297 


•57 


"•4344 


.82 


1.2578 




08 1 


5683 


33 


1.5271 


.58 


1.4292 


.83 


1.2482 




09 1 


5676 


34 


1.5244 


•59 


1.4238 


.84 


1.2381 




10 1 


5668 


35 


1. 5216 


.60 


1.4182 


.85 


1.2277 




11 1 


5660 


36 


1.5187 


.61 


1. 41 25 


.86 


1. 2172 




12 1 


5651 


37 


"■5"56 


.62 


1.4066 


.87 


1.2063 




13 " 


5642 


38 


1.5 1 24 


.63 


1.4006 


.88 


1. 1950 




14 1 


5 6l 3 


39 


1.5092 


.64 


"•3944 


.89 


1.1833 




"5 " 


5620 


40 


1.5058 


.65 


1.3881 


.90 


1.1712 




16 1 


5607 


4i 


1.5024 


.66 


1.3817 


.91 


1.1586 




17 1 


5594 


42 


1.4989 


.67 


"•3753 


.92 


"•"455 




18 1 


5580 


43 


1.4952 


.68 


1.3688 


•93 


1.1318 




19 1 


5565 


44 


1.4918 


.69 


1.3622 


.94 


1 "75 




20 1 


555o 


45 


1. 488 1 


.70 


"•3544 


•95 


1. 1023 




21 1 


5533 


46 


1.4842 


.71 


1.3484 


.96 


1.0860 




22 1 


5515 


47 


1.4802 


.72 


1.3412 


•97 


1.0686 




23 1 


5497 


48 


1.4761 


-73 


*-3337 


.98 


1.0500 




24 1 


5478 


49 


1 .4718 


•74 


1.3261 


•99 


1.0275 














1.00 


1. 0000 



AXES OF ROTATION PARALLEL 1 09 

or the length of the elliptic quadrant is known. We must now find 

what length of major axis, i.e. what value of "a" in conjunction 

with the known value of e will give the required length of quadrant. 

The equation of the ellipse is 

9 1 

9 ■ 79 
or b' 

When written in terms of the eccentric angle <f> (Fig. 78) this 

becomes 

x = a sin <f>, 

y — b COS <f>, 

dx = a cos <fi d(f>, dx 2 = a 2 cos 2 <£ d<f> 2 , 

dy = b sin cf> d<$>, dy 2 = b 2 sin 2 <f> d<f> 2 , 



=f : Vdx 2 + dy 2 , 



= \-y/a 2 cos 2 4> -h b 2 sin 2 <^> d<j>, 

n 

= I -yja 2 — (a 2 — b 2 ) sin 2 <ft d<f>, 

= a l\ 



— sin 2 <5f> d<j>, 



c/0 



e 2 sin 2 </> d<f> = aE. 



The above integral is one of the forms of the " elliptic integral," 
and cannot be expressed in any simpler form. There are, however, 
methods of approximating to its value to any required degree of 
accuracy, and the results will be found in tables of Elliptic Func- 
tions. The accompanying table gives values of E where e advances 

by hundredths, and where c/> is equal to -• 

2 



no 



MACHINERY OF TRANSMISSION 



We see that s can be expressed as 




s = a x E, 




where E is known as soon as e is known. 


But 






hence, a = -> 





4PE 




Fig. 78 



which gives us the semi-axis major of the ellipse which will fulfil 
the requirements. A few problems will place the method in a 
clearer light. 

Let Z— 9, N— 30 teeth, P= 2 (diametral pitch). 

V2+1 3 + 1 
From the table, when e — .5, E = 1.476. Therefore 



^^ 3- J 4i6 x 30 



•o75= 



4 x 2 x 1.476 
b = a Vi - <? = 8.075 V.75 = 6.993, 
£ = 0(1 -<?) = 8.075 --5) = 4-037- 



(See Fig. 79.) 



AXES OF ROTATION PARALLEL 



III 



As another example let Z= 2, N— 30, P= 2, 

V2 — 1 



e = 



.1679. .'. E= 1.5576, 



V2 + 1 

a =f. s 66, £ = 7"-457> r = 6".2 9 8. 

In this case we see that the ellipse is very nearly a circle. 

Sometimes the distance between shafts is given, and we are 




Fig. 79 

required to fill up, by means of elliptic gears, this distance, while 
Z retains some definite value. In this case the shaft distance is 
evidently the major axis, 2 a, which immediately becomes known. 
Then 



P = a 2 {i -e 2 )=a 2 li- 



V* 



4 a 2 Vz 
Vs+il y~(Vs+i) : 
Let the shaft distance be 10", and 2 = 4. 



10 . 20 ^Jz _ 10 x 1.4142 _ „ 



Then a = — = 5, b = 

2 Vz+ 1 



3 .714, 



Vz — 1 1 
e = —r~ x — = - = -3333, 

V2+1 3 

t = a(i-e) = 3 ". 33 . 



112 MACHINERY OF TRANSMISSION 

In this case the circumference becomes fixed, and a given size of 
tooth cannot in general be used. The circumference will be 

4J- = 4<zE = 20X 1-5271 = 3o".542. 

If N= 15 teeth, the circular pitch will be 

i»=^=2".o 3 6. 

Having calculated its dimensions, the ellipse must now be laid 
out on the drawing board. When the major and minor axes 
are given, the best construction is that of Fig. 80. Draw circle 
of radius OB = b, and one of radius OA = a. Draw a num- 
ber of lines through O cutting these circles, such as Ob a. 
Through b draw bP parallel to OX, and through a draw aP par- 
allel to OY. The intersection of these at P is a point on the 
ellipse. We must now lay off the tooth intervals all the way around 
by laying off the circular pitch P'. Bisect each of these intervals 
for the tooth and the space. At the centre of each tooth we must 
now draw a normal to the ellipse. Suppose we wish to draw a 
normal at Q. Draw Qr and Qs. Draw Ors, and produce to / 
where the circle whose radius is {a -f- b) is intersected. Then Qf 
is the required normal. The same is shown at P. When the 
normals at every tooth have been drawn, they will envelop the evo- 
lute of the ellipse, one branch of which is shown at Imn. The point 
of tangency at m between the normal and the evolute will give the 
centre of curvature at P. If required, the radius of curvature can 
be calculated from 

p = r~' 

but this is unnecessary. 

We now construct each tooth as if on a circle whose radius is 
the radius of curvature at the point required. It is evident that all 
the teeth in a quadrant will be different, but if the ellipse is of small 
eccentricity, there will be a number of teeth at the extremities of 
the axes very nearly alike. 

If the number of teeth is odd, the major axis should bisect a 
tooth at one end and a space at the other. If even, the tooth pro- 



AXES OF ROTATION PARALLEL 



113 



file should pass through the extremity of the major axis at each 
end. If these rules are followed, the two wheels will be exactly 
alike, and can be cast from one pattern. Elliptic gears can be cut 
approximately on the milling machine by mounting the blank on a 
reversed trammel. 




Fig. 80 



As has been said, the free foci C and D (Fig. 75) can be con- 
nected by a link in case the shafts overhang, and we need then put 
teeth at the extremities of the major axes only. These will carry 
the wheels past the dead points, and the pull and push of the link 
will carry them the rest of the way, while the pitch ellipses roll 
upon one another. 

Elliptic gears are frequently used for a quick return on small 
shapers, slotting machines, etc. The eccentricity should be less 
than .4 except where a link is used. 
1 



ii4 



MACHINERY OF TRANSMISSION 



3. AXES OF ROTATION INTERSECTING, BEVEL WHEELS 

Thus far we have considered only the transmission of rotation 
by means of gearing between parallel shafts. In this case we have 
seen that the pitch surfaces are cylinders. When the axes inter- 
sect, the cylinders must be replaced by cones having a common 
apex at the point of intersection of the shafts. Such cones will 
transmit motion as pitch surfaces by pure rolling along an ele- 
ment. As in the case 
of cycloidal spur teeth 
where the tooth surface 
is swept up by elements 
of a pair of describing 
cylinders rolling within 
and without the pitch 
cylinders, so in the case 
of cycloidal bevel teeth 
the tooth surfaces are 
swept up by elements 
of a pair of describing 
cones rolling within and 
without the pitch cones, 
all four having a common 
vertex. The intersec- 
tion of the tooth surface 
with a sphere whose cen- 
tre is the common apex of 
the cones forms a curve 
in space known as the 
spherical epi- or hypo- 
cycloid, PX, FY (Fig. 81). The involute bevel tooth surface is 
swept up by a line of a plane which rolls between base cones, coaxial 
with but with smaller vertical angle than the pitch cones. The 
generating line must be one which always passes through the apex 
of the cones. Fig. 82 shows the generation of the spherical invo- 
lute. The plane GEH (taken for clearness as a circle with radius 




Fig. 81 



AXES OF ROTATION INTERSECTING, BEVEL WHEELS 115 

equal to the slant height of the cone) rolls upon the base cone 
DA C, AE being the line of tangency. Any radius of the circular 
plane such as AX will sweep up the tooth surface, the point X 
sweeping up the spherical involute XP. Thus we see that the 
bevel tooth curves may be treated by spherical geometry just as 
spur tooth curves were treated by plane geometry. Small circles of 




Fig. 



the circumscribing sphere will replace circles in the plane geometry, 
and great circles of the sphere will replace straight lines. Thus 
the spherical epicycloid is traced by a point in the circumference 
of a small circle rolling outside of another small circle. The spheri- 
cal cycloid is traced by a point of a small circle rolling upon a great 
circle. This would be the tooth curve of the crown wheel or bevel 



n6 



MACHINERY OF TRANSMISSION 



rack. The spherical involute is traced by a point of a great circle 
which rolls upon a small circle, etc. Being on the surface of 

a sphere, the spheri- 
cal involute does not 
pass to infinity, but 
terminates in a se- 
ries of cusps upon 
the base cone and its 
prolongation. (See 
Fig. 83.) 

For drawing bevel 
gears, Tredgold's ap- 
proximation is used 
(Fig. 84). At the 
base of the pitch 
cone another is 
drawn, having its ele- 
ments at right angles 
to those of the pitch 
cone. This " normal 
cone" is developed 
upon a plane, and 
the teeth are laid 
out upon its devel- 
oped circumference as on a spur gear. This is then wrapped 
back upon the normal cone, and the tooth surface is supposed to 
be generated by a line which always passes through the vertex of 
the pitch cone, and always touches the tooth profile. In practice 
only frusta of the pitch and normal cones are used, as shown in 
Fig. 84. The slant heights of the frusta should seldom be more 
than one-third those of the cones. 

Bevel wheels whose shaft angle is 90 are called Mitre Wheels. 

Since all the surfaces of bevel teeth are conical by reason of their 

generation, they cannot be cut correctly with a rotary cutter, as such 

a cutter can make a cylindric surface only. However, they can 

be cut approximately by following the rules given in Brown and 




Fig. 83 



AXES OF ROTATION CROSSING, SKEW WHEELS 117 

Sharpe's " Formulas in Gearing," published by the company.* 
The teeth can be cut perfectly only by planing them out element 
by element on a special machine. With such a machine twisted 




Fig. 84 



bevels can be cut by giving the blank a reciprocating circular 
motion in unison with the reciprocating rectilinear motion of the 
cutter. 

4. AXES OF ROTATION CROSSING, SKEW WHEELS 

A. Spiral Gears 

We have seen how a twisted gear is derived from a spur gear, 
and works according to exactly the same theory. Suppose CD 
(Fig. 85) to be an ordinary twisted gear, in mesh with a piece of 
twisted or oblique rack EFGH. If CD is turned with a constant 
angular velocity counter-clockwise as viewed from A\ EF will 
move with a constant velocity PV in the direction EF, and there 
will be no sliding parallel to the elements of the rack teeth. But 
if the rack is constrained by means of guides to move in any other 



* For the dimensions of bevel teeth consult this book also. 



u8 



MACHINERY OF TRANSMISSION 



direction, the gear will drive it at some other constant velocity, 
and there will be a sliding parallel to the elements of the rack teeth. 
In fact, it is self-evident that a twisted gear will drive an oblique 

rack in any direc- 
tion except paral- 
lel to its own teeth, 
and that the ratio 
between the ve- 
locity of the pitch 
surface of the gear 
and that of the 
rack will be a con- 
stant. 

Now suppose we 
have instead of a 
solid rack merely 
the surface of a 
rack, such as might 
be made by bend- 
ing a thin sheet of 
metal into an ob- 
lique rack. Then 
a second twisted 
gear could mesh 
with this rack from 
below, and if the rack were driven in any direction by the upper 
gear, the two gears would turn with a constant velocity ratio. In 
Fig. 86 let the twisted gear EFGH mesh with a piece of rack, 
as shown from above, and let this mesh with another, ABCD, 
below. We will suppose for simplicity that the teeth are involute. 
The lines RS and TU are the lines of the root edge and the tip 
edge of a rack tooth when seen from above. This is shown in per- 
spective in Fig. 87. To the right, and above in Fig. 86, are shown 
projections of the faces EF, GH, AB, and CD of the gears and 
rack, showing the tooth faces of the latter by the lines e'f', g'k', etc. 
The point of contact on the face EF will be where the line of 




AXES OF ROTATION CROSSING, SKEW WHEELS I 1 9 

obliquity intersects at right angles the tooth outline at m\ and when 
projected back gives the point m in its proper position. In the 
same way we find for the face GH the tooth face g'/i', the point 
of contact /', and the projected point /. Now when viewed from 
below the line ftS becomes the tip edge, and the line TU the 
root edge. Projecting the faces AB and CD of the lower gear 




upward, drawing the obliquity, and projecting the contact points 
back, we find the point n of contact on the face AB and the point 
o of contact on the face CD. Hence the upper gear touches the 
rack along the line ml, and the lower gear along the line no. These 
lines are evidently the oblique generating lines PQ (Fig. 71) of 
the teeth of the two wheels, which in twisted gears are coincident. 



120 



MACHINERY OF TRANSMISSION 



If now we remove the rack surface, the two gears themselves will 
touch in a single point E, which is the intersection of the two lines. 
Hence spiral gears, which are nothing more than twisted gears 
with shafts askew, will touch in one point only, but they will main- 
tain a constant angular velocity ratio, since the lines ml and no will 
continue to intersect at some point as the action continues. Two 
twisted gears will therefore work together at any angle, but the 
wheels must be at the shortest distance between the crossing shafts. 
They can be cut on a milling machine by giving the blank a motion 
of rotation about its own axis as it passes under the cutter, and the 
tooth surface is approximated by the ordinary spur tooth cutter. 




/\ 5 



The true circular pitch of a spiral gear is the pitch circumference 
divided by the number of teeth, viz., CD (Fig. 72). The normal 
circular pitch is the perpendicular distance between similar ele- 
ments of two adjacent teeth measured upon the pitch surface, or 
CE. This will be the circular pitch of the rotary cutter. Hence 
we cannot calculate the pitch of the cutter in the ordinary' way. 
If we call the angle between a tangent to a tooth on the pitch sur- 
face and a line parallel to the axis, the spiral angle, and denote 
this by a, then 

CE = CD cos a. 

If E' is the true circular, and E n ' the normal circular pitch, 

2 ^ R = P> 
N 



AXES OF ROTATION CROSSING, SKEW WHEELS 121 
exactly as in the case of spur gears ; but 



2 7r R cos a 



N 



P' 



and 



PS 



where P is the diametral pitch of the rotary cutter. Hence, 

p= f ■ 

2 it cos a 

We will first deduce the relation between the angular velocity 
ratio and other constants of the gear. Consider two spiral gears 
whose axes are A A' and BB' (Fig. 88). Let XY be a common 




element at the point of contact O. Let XY move through a 
differential distance to X'Y'. Then a point of wheel A has 
moved through a distance ^ and a point of i? through a distance 



122 MACHINERY OF TRANSMISSION 

V 2 . If we call the normal distance between XY and X'Y' V n ,we 

will have 

V 1 cos a = V n — V 2 cos /?, 

V Y _ cos ft 
V 2 cos a 

Now V 1 and F 2 may be considered the velocities of the circum- 
ferences of the two gears, and if oi x and w 2 are their angular 
velocities, and R l and R 2 their pitch radii, 

wj _ V X R 2 _ R 2 cos /? 
w 2 V 2 R Y R 1 cos a' 

or the angular velocity ratio is equal to the inverse ratio of the 
radii multiplied by the inverse ratio of the cosines of the spiral 
angles. 

Suppose in a pair of spiral gears we are given the value of 8, or 
the inclination between the shafts, the angular velocity ratio 

— = 7}, and the ratio of the radii — ■ . We are also given N u the 
o) 2 R 2 

number of teeth on one wheel, and P, the diametral pitch of the 

rotary cutter. Then as 

<* + £ = S, 

<!>! R 2 cos(8 — a) R 2 (cos 8 cos a + sin 8 sin a) 
w 2 R ± cos a R x cos a 

n 

rj — = cos 8 + sin 8 tan a, 

or a = tan -1 <{ -^ -~ — cot 8 \ • 

{ R 2 sm 8 J 

Then we know j3 from 

(3 = 8- a. 

Now />, the diametral pitch of the rotary cutter, is the same for 
both wheels, as is — = P n ', the normal circular pitch also. Then 

P' = PJ = true circular pitch of R lf 

cos a 

and P 2 = PJ ^ = true circular pitch of R 2 ; 

cos/3 



AXES OF ROTATION CROSSING, SKEW WHEELS 123 



hence, 



R 1 = 



R,= 



PJN X 

27T 

PJN* 

2 IT 



#1 



2 P cos a 

No 



2 P cos (3 

The dimensions of the blanks and the cutting angles being now 
determined, we must select the proper cutter of the set whose 
pitch is given. If we select any tooth upon a spiral gear and pass a 
plane through it normal to its helical elements, the intersection of 
this plane with the pitch cylinder will be an ellipse whose minor axis 



is 2R, and whose major axis is 



2R 
cos a 



, a being the spiral angle. 



The spur gear whose tooth would most nearly coincide with the 
profile of the spiral gear selected would be the gear cut upon the 
osculating circle at the extremity of the minor axis of the ellipse. 
The radius of curvature at the point mentioned is found in the 
ordinary way as follows : the equation of the ellipse is 



I 
b 2 



h 



dx 



bx 



:Vfl s 



d*y 

die 2 



= T 



ab 



[ 



(a 2 -x 2 ) 
dy\ 



1 + 



dxj 



dx' 



1 + 



b 2 x 2 



a 2 (a 2 -x 2 ) 



ab 



{a 2 -x 2 )? 
At the extremity of the minor axis x = o, so we have 



1 

ab 



124 MACHINERY OF TRANSMISSION 

Now in the ellipse formed by the intersection above described 

a = , and b = R. Hence, 

cos a 

R 2 R 

Po 



R cos a cos 2 a 

This is the radius of curvature of the " Osculating Spur Gear." 
The number of teeth on this gear would be 

2 7iy> 2 ttR 









■""O — 


R' ~ 


Pn 


cos 2 a 


But on 


the 


spiral 


gear itself we have 










N — 


2ttR 


2 irR cos a 






P 




P' 








Nn — 


N 








J-Vq -— 


cos 3 a 





The cutter should therefore be chosen from the set, not according 
to the number of teeth on the spiral gear itself, but according to 
the number on the osculating spur gear. 

If 8 = 90 , the following special cases may occur : 





a = tan -1 — - rj. 
P2 


If in addition R x = R 2 , 


a = tan -1 rj. 


h,=i, 


a = tan -1 — • 
P2 


If R 1 — R 2 , and 17 = 1, 


a = tan -1 1 = 45 






and the two wheels will be exactly alike. 
If 8 = o, the case reduces to a spur gear. 

Critical Angles or Angles of Maximum Efficiency. — It has now 
been shown that the fundamental laws connecting the various 
quantities in the spiral gear are 

(Oi_ A^cosJ? , v 

w 2 R\ cos a 

X=R l +R 2} . . . . . (2) 
8=a + /3, . . . , . . (3) 



AXES OF ROTATION CROSSING, SKEW WHEELS 125 

where K is the distance between shafts. The three quantities on 
the left sides of the equations are evidently fixed once for all by 
the requirements of the design, leaving four unknown quantities, 
viz., «, (3, fi^ and R 2 , one of which must be chosen arbitrarily, or 
by some other independent condition. This condition may be 
made that of maximum efficiency. 

Loss of power on spiral gears is due to friction (1) in the bear- 
ings, and (2) between the teeth. The first of these can be further 
separated into (a) journal friction due to the obliquity of action 
of the teeth forcing the shafts apart, as in any form of gearing, 
and (£) end-thrust friction. The sliding between the teeth may 
be separated into (a) a. component of approaching and receding 
action, also common to all forms of gearing, and (b) a component 
of sliding along the helical elements of the teeth. Now it is useless 
to attempt to deduce that diameter of gear and angle of tooth 
which, with given coefficients of friction, would reduce all these 
various losses to their combined minimum value, for even when 
obtained the result would be too complicated and uncertain to be 
of any practical value. So let us see which of the above are the 
most wasteful of energy which are small enough to be neglected, 
and which may be omitted because their effects would not be 
greatly changed by change in radius and angle. The two losses 
designated (a) will be small compared with {b), particularly in 
the first case. Roughly speaking, their effects will be smallest 
in wheels of equal size, though change in radii and tooth angles 
will not greatly vary their magnitude. Next in importance 
will be the end-thrust factor, which may be varied widely by 
change in the angle of the tooth. And last and most impor- 
tant of all is the sliding between the teeth along the helical 
element. 

Omitting first all other sources of loss of power, let us investigate 
that of end thrust. Let A (Fig. 89) be the driver through which 
power is transmitted to B. Let ON be the normal pressure or 
force active between the elements of the teeth in the tangent plane 
to the two wheels. Then OP, the projection of ON in the plane 
of the gear A, will \)$ the effective driving force of A in its. own 



126 



MACHINERY OF TRANSMISSION 



plane, and NPw'iW be the end thrust on its shaft. Similarly OQ 
will be the driving force on B, and NQ its end thrust. 

Now the total power 
given to the system 
by A will be equal to 
{OP) x V 1} where V 1 
is the velocity of the 
surface of A in the 
direction of OP. The 
power lost due to fric- 
tion through the end 
thrust of A will be 

(PN) x v' X <f>, 

where v' is the mean 
Fig. 89 velocity of the end- 

thrust collar, and <£ 
the coefficient of friction at that point. Hence the ratio of power 
lost to total power in the system due to this single cause will be 

£ = PNxv'x<f> + QNxv" Xcj> 
OPxV x 

But v' = r^, and v" = r x w 2 , where r x is the mean effective frictional 
radius of the end-thrust collar taken the same for both shafts ; also 
PN= OP tan a, QN= OQ tan /?, and V 1 = P 1 w 1 , which substi- 
tuted give 

P _ ^(£((0! X OP tan a -f- <o 2 x O Q tan /?) 
OP X iVi 

Furthermore, 0Q= OP — -", so that the equation reduces to 




cos a 



E=r l 4>\ 



f 0^ sin a -f- w 2 sin /3 



^wj cos a 



(4) 



In this we must substitute for all unknown quantities in terms of a, 
and find what value of a will make E a minimum. Combining 
equation (1) with (2), and solving for R x , we get 



AXES OF ROTATION CROSSING, SKEW WHEELS 1 27 
D X(i)<> cos 3 

Jv 1 = — ) 

w 1 cos a + w 2 cos /? 

and substitution gives 

P _ r^ \ (cox sin « + ^2 sm /?)( M i CQS « + <^2 cos P) \ 
K [ o> 1 (o 2 cos a cos /3 J 

which by reduction becomes 

r, 1\<I> f cui sin « . o)o sin B , , , . «_] , >. 

^ = -^i — ^ + — ^- + tan«-f-tan/3 !-. . (5) 

K [ 0> 2 COS /5 <u 2 COS « J 

That value of a in the general case given above which makes E. a 
minimum leads to an unsatisfactory result when the shaft angle 8 

and the angular velocity ratio — , which we may call rj, are small, 

0)2 

for then there may be no real minimum value of a within the 

angle 8. But fortunately in the particular case where the shafts 

cross at 90 , the solution is easily applicable to all cases. In this 

case 

^^^(ry + I + tanft + tan 
A l 77 

Before applying the ordinary methods of finding that value of 
a which makes E x a minimum, we must notice a certain restric- 
tion which must be placed on all the above equations. Should for 
any reason the algebraic sign of either of the angles a or /? change, 
the sign of the resulting power lost will not change. In other 
words, the formulae do not necessarily hold for any other than 
positive angles, or angles between the limits of 8. Differentiating 
the above equation with respect to a, and putting the first differ- 
ential coefficient equal to zero, 

dE 1\d» co o n ) 

— = -=- 1 sec- a — sec 2 (3 \ = o, 
da K 

sec 2 a = sec 2 /?, 

a=p. 

For positive values of a and (3, this is easily distinguished as a 
minimum, and the result is independent of r 3 <j>, or K. In this 



128 MACHINERY OF TRANSMISSION 

case then it would seem that teeth set at an angle of .45° would 
give a minimum of work lost by end thrust. 

In the problem of minimum power lost by the sliding of the 
teeth along a helical element, the angle a has usually been taken 
as that angle giving the minimum velocity of slip in that direction, 
with given fixed angular velocities of both shafts. Referring again 
to Fig. 89, if OC=V 1 is the velocity of a point on the pitch 
surface of A in its own plane, and tangent to its surface, and 
OD = V 2 the velocity of a point similarly related to B, the length 
CD will be the velocity of these points relatively to one another 
along the tangent to the helical element. Calling this velocity 
of slip v 8i then 

v s = V 1 sin a-\-V 2 sin /3, 

= R x <*>! sin a-\- R 2 oi 2 sin f3, 

= R Y (o x sin a -f (K — R x ) sin j3. 

But from the previous case 

„ _ Ka) 2 cos (3 



wj cos a + w 2 cos fi 
Hence by substitution 

v s = ^^ ] cos a sin j3 + cos /5 sin a 

toj cos a -f w 2 cos /5 l 

Koy^.2 sin 8 
w 1 cos a + <*>2 cos ft 

Now the numerator of this fraction is a constant, hence v s will 
vary only by the variation of the denominator, or v s will be a 
minimum when (a^ cos a + a> 2 cos/3) is a maximum. Call this 
quantity Y, then 

dY . 

= — 0^ sin a -+- o>2 sin p = o, 

da 

wj _ sin (3 

oo 2 sin a 

By using again the positive pair of angles given by this last we can 
easily distinguish Y as a maximum. If along the axis of A (Fig. 
90) we lay off the angular velocity ^ of A to any scale as OL, 
and along B lay off w 2 equal to OM, and complete the parallelo- 



AXES OF ROTATION CROSSING, SKEW WHEELS 



29 




Fig, 90 



gram OLFM, the diagonal OF will give the proper values of a 
and p to give minimum velocity of slip. It will be noted that 
since the ratio of angu- 
lar velocities is equal to 
the inverse ratio of the 
numbers of teeth, we 
might lay off the num- 
ber of teeth of B along 
the axis of A, and vice 
versa. This same re- 
sult has been shown by 
MacCord and others by 
different methods of 
procedure. 

But it does not appear 
that the angles giving 
the minimum velocity 

of slip are necessarily those giving the minimum power lost by 
that slip. Power is composed of two factors, force and velocity, 
and it is the product of these and not one alone which must be 
a minimum. Attacking the problem then along the lines of the 
first one, we see that 

Power put in by A — FV 1} 
Power lost due to v s = N$v„ 

where N is the normal pressure between the teeth, and <j> the co- 
efficient of friction. 

Then it is E = W»*=#&±, 

FV X FR^ 

which must be a minimum. From the figure, 

P 

cos a 

Kui^o sin 8 



N: 



From the preceding, 



md 



v s = 



*1 



(o 1 cos a -f- <o 2 cos (3 

Koio cos ft 
w 1 cos a 4- «2 c °s j8 



130 MACHINERY OF TRANSMISSION 

Substituting these in the expression for E, it reduces to the 
simple form 

<£ sin 8 

±L = -> 

cos a cos p 

and rejecting constants, 

cos a cos p 

is to be a minimum, or F= cos a cos (S — a) 
is to be a maximum. Differentiating, 

— = — sin a cos (3 -f sin (3 cos « = o, 
sin (ft — a) = o, 

By the ordinary methods of second differentiation, this may be 
shown to be a maximum, and therefore when the tooth bisects the 
shaft angle, the power lost by sliding between the teeth as above 
stated is a minimum. 

When a and ft are known, i? x is found from 

_ Xcos/3 
1 77 cos a -j- cos /? 

and R 2 = K- R x . 

The methods of using the above formulae can best be seen by 
applying them to a definite case. Let the distance between the 
shafts be K = 6", and let the shafts cross at 90 . The angular 

velocity ratio is to be rj = — = -J-, and P= 10. The angles are to 
be those giving the maximum efficiency. Then 

a = P= 45°, 

R Xcos/3 = K 

1 77 cos a + cos (3 r] + * 

R 2 = £-R 1 =2". 

The numbers of teeth on the two wheels must now be calculated. 
These in general will not be whole numbers, and as a fractional 




A Yli- 73 I A .5 Cos (£ .6 I J HS I .9 | fi.O 

80° 70° 60° tC 50° 40° 30° 20° 10°0° 



2PK 



Cutting Angles of Spiral Gears. 



AXES OF ROTATION CROSSING, SKEW WHEELS 131 

tooth is an impossibility, we must select the nearest whole numbers 
to the ones found, and recalculate the angles and radii to fit the 
new case. The numbers of teeth will be found from 

N-i = 2 PR X cos a, 

N 2 =2 PR 2 cos p, 

which in the above example become 

JVi= 2 x 10 x 4 X .7071 = 56.568, 

and N 2 = 2 x 10 x 2 x .7071 = 28.284. 

Hence we must select for our gears ^ = 56 teeth, and JV 2 = 28 
teeth, as being the numbers giving the highest efficiency. Trans- 
posing the equations for tooth numbers we have 



£1 = 

and P 2 = 



2 P cos a 



2 P cos p 2 P sin a 



Hence, K — — = h 



2 P cos a 2 P sin a 



t a 2 ' 8 . T '4 
In our case 6 = h 



cos a sin a 

If we could solve the above equation for a, our inverse compu- 
tation would be complete. But unfortunately the equation is of 
the fourth degree, and, though possible of solution, such solution is 
not practical. Furthermore there are four real values of a which 
will satisfy it. Graphic methods or continued approximations must 
then be resorted to. One of the best graphic methods appears to 
be that of Robert Bruce,* which is as follows : let OX, OY (Fig. 
91) be a pair of rectangular coordinates. Lay off upon the axis 

of X — ^, and upon the axis of Y -~, thus determining the point 
* American Machinist, April 12, 1900. 



132 



MACHINERY OF TRANSMISSION 



C. Lay a graduated scale upon the paper, so that its edge passes 
through C, its zero point lies upon one of the axes, and shift it 
until it intercepts K inches between the axes. Then the angles a 
and fi in the figure are evidently the required ones. The appended 
Diagram No. 3 gives a method by which the angles can be read 
off directly. Having obtained the nearest whole numbers of teeth 
on the gears, find on the diagram the point G, whose coordinates 




^1 A 

are — =rr; and 



N t 



Fig. 91 

..., on the inner scales. Through this point 
2 l^K 2 r^K 

draw a line or merely lay a straight edge tangent to the curve 

representing the shaft angle. The outer scales on the bottom and 

left will give roughly the angles a and /? respectively, and the inner 

scales the values of cos a and cos (3 quite accurately. The radial 

lines of velocity ratio will facilitate the locating of the desired point, 



AXES OF ROTATION CROSSING, SKEW WHEELS 133 

for if the ratio be one of those given, the point must lie on 
its line. 

It is interesting to note that two lines can be drawn through a 
given point tangent to the curves as shown. As a matter of fact, 
four such lines could be drawn provided the whole of the curves 
were drawn in, but that portion shown is the only portion giving 
positive angles, i.e. angles within the angle 8. But there will be 
two separate positive values of the angle a, which, with a given 
velocity ratio, number of teeth, and shaft distance, will work 
correctly together, giving of course different values of the radii. 
Which of the two is the one required can be easily told as lying 
nearest to the first approximation of the angle. The same result 
is seen in the case of Mr. Bruce's solution. Two positions of the 
line AE (Fig. 91) can be found passing through C, where the 
length is K inches. If the point G lies on one of the curves, 
the two positions coincide, a limiting case, and if it lies on the 
concave side, the solution is impossible within the angle 8. 

By the application of either method to our problem, we find 

« = 46° 33', 
P = 43° 27', 

*i = -#— = —& = 4".o 7 2, 

2 P cos a 20 x cos a 

j? N 2 28 Q 

J\-2 Z=Z = = I.928. 

2 P cos (3 20 x cos (3 
The cutters are then selected from 

N ' = -^ = 169 = No. 1 (rack), 
cos 3 a J 

N »'=-^h= 73 = No. 2. 

cos 3 fit 
As a more general case take the following : let 8 = 6o°, K— 6", 
rj = — = ^-, P= 10. Suppose certain considerations in the design 

(U 2 



134 MACHINERY OF TRANSMISSION 

make it necessary that R x should be as nearly as possible 4! inches, 
then approximately R 2 = if. 

a=tan- 1 {— 1 ^--cot8J, 
I R 2 sin 8 J ' 

= tan" 1 { 4i^5 _^5 _ cot 6o o ] = o / 
J 1.75 sin 6o° f ^ 6 > 

Q = 8 — a = 6o° — 39 31'= 20 29', 
iVi = 2 7^ cos a = 65.572, 
N 2 =2 PR 2 cos /? = 32.787. 
In this case we would have to use 

N x = 66 teeth, 
and N 2 = 33 teeth. 

From the diagram, or by repeated trials from 





3-3 j i-65 


= 6. 






cos a cos (8— 


a) 




or 


2 sec a + sec 


P=3- 


6364, 


we find that 


a = 3 S c 


43', 






0=21 


' 17', 




from which the exact radii are 








*i = 4" 


.229, 




and 


tf 2 =i" 


771. 





Method of cutting Spiral Gears. — The lead of a helix or screw 
is the distance between two of its consecutive intersections with 
an element of its pitch cylinder. If the helix is developed by 

C 

rolling on a plane, it is readily seen that — = tan a, where C is 

the circumference of the pitch cylinder, / the lead, and a the spiral 
angle of the helix. In cutting a spiral gear on a milling machine, 
the blank is first set at the proper spiral angle under the rotary 



AXES OF ROTATION CROSSING, SKEW WHEELS 1 35 

cutter. It is moved forward by means of the screw, and also 
rotated by means of the spiral head. It is evident that when the 
screw has moved it forward a distance equal to the lead, the spiral 
head must have rotated it through one complete turn. The screw 
is connected to the worm which actuates the spiral head by a train 
of four gears. The numbers of teeth on these gears we will repre- 
sent by a, b, c, and d. If there are N revolutions of the screw 
corresponding to n revolutions of the worm, then 

N ~ = n. 
ac 

Now it requires 40 revolutions of the worm to rotate the spiral 
head once, and there are 4 threads per inch on the screws of the 
Brown and Sharpe milling machines. Hence to move the blank a 
distance equal to the lead requires, N= 4/ turns, and to rotate the 
spiral head once requires n = 40 turns, or 

hd 
aI— — 40. 
ca 

, 2ttR 

But /= 



tan a ' 

2ttR bd 

hence, X — =10, 

tan a ca 

R and a being given, the ratio — must be so chosen as to satisfy 

the equation. In the directions which come with the machines, a 
number of such combinations are worked out. It is evident that 
but few angles can be cut with absolute accuracy. (1st gear on 
stud=^ teeth, 2d gear on stud = c teeth. Gear on worm = a 
teeth, gear on screw = d teeth.) 



B. Worm Gears 

If the spiral angle of one gear is very nearly 90 , one tooth may 
be made to return on itself, and the wheel reduces to an ordinary 
screw. If in addition the angle a is nearly 90 , the spiral angle of 
the other wheel is very small, and is in fact very nearly a spur gear. 



136 



MACHINERY OF TRANSMISSION 



This combination is called a Worm Gear. In the case of the 

worm, since p = — %— , when a approaches 90 , p approaches 00, 

or the tooth outline of a worm is practically that of a rack. 

The action of a screw upon a gear can be much improved by 
the following method : a copy of the screw is made in tool steel, 
and this is notched parallel to its axis like a tap. The wheel is 
notched into the required number of teeth upon a milling machine. 
The notched screw or " hob " is rotated between centres, and the 
wheel, free to revolve upon a stud, is fed up against it. The 
cutting edges of the hob soon work into the notches of the wheel 
blank, until the proper depth is reached. The hob is then re- 
placed by the original screw. Such concave gearing is excellent 
for causing slow or small amounts of motion, but is not very 
economical for transmitting power, as the sliding between the teeth 
is large. A worm wheel will not work properly if it contains less 
than from 25 to 30 teeth. 



TABLE FOR DIAMETRAL PITCH OF WORM TOOLS. (Grant)* 



Diametral Pitch 


1 


2 


3 


4 


5 


6 


7 


8 


Point of Hob tool . 


1-035 


.517 


•345 


.258 


.207 


•I7S 


.148 


.129 


Point of Worm tool 


.968 


.484 


•323 


.242 


.197 


.162 


.138 


.121 


Depth of Cut . . 


2.125 


1.063 


.708 


•532 


425 


•354 


.304 


.266 


Increase . 


.250 


.125 


.083 


.063 


.050 


.042 


.036 


.032 


Diametral Pitch 


9 


10 


11 


12 


13 


14 


15 


Point of Hob tool . 




•"5 


.104 


.094 


.086 


.078 


.074 


.069 


Point of Worm tool 


.105 


.097 


.088 
•193 


.081 


•073 


.069 


.064 


Depth of Cut . . . 


.236 


.213 


.177 


.164 


.152 


.142 


Increase 


.028 


.025 


.023 


.021 


.019 


.018 


.OI7 



* " The Teeth of Gears," by George B. Grant, Lexington Gear Works. 



AXES OF ROTATION CROSSING, SKEW WHEELS 137 
TABLE FOR CIRCULAR PITCH OF WORM TOOLS. (Grant) 



Circular Pitch 


2 


if 


ii 


i? 


Ig 


1 


§ 


3 

5 


Point of Hob tool . 


.644 


•5 6 4 


483 


.402 


.362 


.322 


.282 


.241 


Point of Worm tool 


.620 


.542 


.466 


.388 


•349 


.310 


.271 


.233 


Depth of Cut . . 


1.416 


1.240 


1.062 


.886 


•797 


.708 


.620 


•53i 


Increase . . . 


.166 


.146 


•125 


.104 


.094 


.083 


•073 


.062 


Circular Pitch 


1 


h 


XB 


3 


ft 


1 
3 


& 


Point of Hob tool . 


• • 


.201 


.161 


.141 


.121 


.100 


.080 


.060 


Point of Worm tool 


.194 


•155 


•135 


.116 


.097 


.078 


.058 


Depth of Cut . . 


•443 


•354 


.3IO 


.265 


.222 


.177 


•133 


Increase 


.052 


.042 


.O36 


.031 


.026 


.021 


.016 



" The sides of the tool should come together at an angle of 30 . 
Make the tool the proper width at the point, and thread the 
worm to the required depth of cut. Make the diameter of the hob 
greater than that of the worm by the amount of the increase. 
Grind off half of the increase from the point of the tool, and use it 
to thread the hob to the same depth of cut." — (Grant.) 



C. Hyperboloidal Gears 

The gears we have just been considering have but one point of 
contact between the teeth. Gears can be constructed, however, 
upon shafts which cross without intersecting, having a true line 
contact. Nor must these wheels be placed at the shortest distance 
between shafts, but may be anywhere along the axes. 

The pitch surfaces are hyperboloids of revolution of one sheet, 
for let A A' and BB' (Fig. 92) be two axes, crossing without inter- 
secting, their shortest distance apart being at O. From the defini- 
tion of pitch surfaces they must touch along a line, such as XY. 
Then as the axes revolve, the line XY will sweep up an hyper- 



138 



MACHINERY OF TRANSMISSION 



boloid with respect to each axis. Hence the line is called the 
Generatrix. If viewed directly from above as in the figure, a and 
(3 will be the angles between the generatrix and the axes. The 
surfaces are not necessarily tangent along XY unless the proper 
relation exists between «, (3, R x , and R 2 . If they are tangent at 
any point, there must be a common normal to the surfaces at that 
point. But these surfaces are surfaces of revolution, and every 
normal to such a surface intersects its axis. Hence if two hyper- 
boloids are tangent along the generatrix, a perpendicular let fall 




Fig. 92 



from one axis upon the generatrix will, if produced, intersect the 
other axis. The line connecting the shortest distance between 
shafts is such a line, and therefore the generatrix must pass through 
the point O in the figure. Let PSQ be another such line. ' Since 
the axis and generatrix are both parallel to the plane of projection, 
the projection of the angle PSO is a right angle as shown. 



Hence 



QS _ tan a 

PS ~ tail 

Fig. 93 is a horizontal projection of the same two hyperboloids, 
and the points P', <7, and S' are found by projecting P, S, and Q 
of Fig. 92. Now if a straight line be divided into any two parts, 



AXES OF ROTATION CROSSING, SKEW WHEELS 



139 



the ratio of these will be equal to the ratio of their projections 
upon any plane, therefore 



RS 
QS 



/T7- \ R'S' /T7 . v tan a 

~ ~7^, ( Fl S- 93)=— g- 



Q'S' 
R'S' = R, 

q's' r 2 ' 

tan a _ R x 
tan J3 ~ A^' 

It is evident that if the tooth surfaces are so formed that their 
intersections with the pitch hyperboloids are coincident with the 



But 



Hence 



a' 1 














^^ f r t ( 


Q'\ A 


V4'k V'Y 


\Js' T 


B 




r 











\p' B 





























Fig. 93 

generatrices, a and /? will be the spiral angles of the two gears. 
There will be positive driving at right angles to the generatrices, 
and sliding along them. As in the case of spiral gears (Fig. 88), 
we haye 

o)i _ R± cos (3 t 

o>2 R x cos a 



but 



Hence 



R 2 _ tan /3 
Ri tan a 

co! _ sin /? 
o)o sin a 



140 



MACHINERY OF TRANSMISSION 



This is the same relation that we obtained when investigating the 
angles that give the least velocity of slip for spiral gears. (Com- 
pare Fig. 90.) 

The method of producing cycloidal skew teeth, analogous to 
that used for spur and bevel teeth, fails in the present instance, 
for it has been shown by MacCord and others that tooth surfaces 
swept up by the generatrix of a small describing hyperboloid rolling 
within and without a pair of pitch hyperboloids will intersect 
along the generatrix instead of being tangent along it. (See also 
the American Machinist, September 5, 1889.) 




Fig. 94 



Involute skew teeth can, however, be formed by a peculiar 
method. Let OA and OB (Fig. 94) be the axes of the pitch 
hyperboloids, tangent along the line OX, and with gorge circles 
tangent at O. Construct two cylinders, coaxial with the hyper- 
boloids, and of diameters equal to the diameters of the gorge 
circles. These cylinders will be tangent at a single point, O. If 
we place a plane between the cylinders in such a way that it is 
tangent to both, this plane will contain the generatrix OX. If the 
plane be moved in the direction of the arrow at right angles to 
the generatrix, it will rotate the cylinders as if by friction, but 



AXES OF ROTATION CROSSING, SKEW WHEELS 141 

there will be a sliding between the plane and the cylinders along 
the elements of the cylinders. If the generatrix is carried along 
with the plane it will sweep up a surface with respect to each 
cylinder, which is known as an Involute Spiraloid. Such surfaces 
will work together as perfect tooth surfaces if we make them 
of proper length to avoid interference. They will always be tan- 
gent along a line which is the generatrix itself. 

The gears made by O. J. Beale for the Brown and Sharpe Mfg. 
Co. are made according to the above theory, which is due to 
Theodore Olivier. Beale's gears are planed out, the straight cut- 
ting edge of the tool acting as the generatrix. 

Approximate skew gears can be drawn by Tredgold's approxi- 
mation in the same way as bevel gears. 



CHAPTER V 

TRANSMISSION OF RECTILINEAR TRANSLATION 

I. PRISMATIC GUIDES 

If a point in a given piece of machinery is to be moved in a 
straight line, its motion may be constrained by means of guides. 
The form of these sliding pairs may be either cylindric or pris- 
matic. The former is used usually where a rod works through 
packing. Prismatic guides are used in nearly all other cases, the 
flat guide to take up heavy pressures, and the V-shaped to prevent 
lateral motion. The problem of the prismatic guide is purely one 
of machine design. 

2. PARALLEL MOTIONS 

A. Classification 

When a point is guided in a straight line, either wholly or in 
part by an assemblage of turning pairs, the mechanism is called a 
Parallel or Straight Line Motion. All forms of parallel motions 
will fall under one of the following four kinds : 

f i. Composed wholly of turning pairs. 
[2. Composed of turning and sliding pairs. 

(3. Composed wholly of turning pairs. 
[4. Composed of turning and sliding pairs. 

B. The Cycloidal Straight Line Motions 

If a circle rolls within another of double its diameter, every point 
on the circumference of the inner circle describes a straight line. 
This can be used as a parallel motion by constraining the centre 
of the small circle to move in a circle by means of the link OC 

142 



PARALLEL MOTIONS 



143 



(Fig. 95). The point P will then move in a straight line A OB. 
But as two circles can be forced to roll together by means of gear- 
ing only, this form belongs to 
Class 2. 

In the case of the epicy- 
cloid, the rolling wheel turns 
in the same direction as the 
revolution of its centre. But 
by the introduction of an idle 
wheel, this can be reversed. 
At the centre O of the fixed 
gear (Fig. 96) an arm is hinged 
upon which are pivoted two 
gears, D and C, C being half 
the diameter of O. Upon C 
is fastened a link CP equal 
in length to CO. Then the 
triangle PCO is isosceles, and angle OPC 
Let angle PCO = fi. Then 

l 6 = 7r-2y. 

Let CO revolve through an angle a. Then wheel C has turned 
through 2 a, and S = ($ + 2 a, or 




angle PO C = y. 



So we have 



B = tt — 2 y -f- 2 cc. 







1(tt — 8) —\ (tt — tt + 2 y — 2 a) 



But angle POC' = y 



y -a = Angle P'OC. 

Thus OP coincides with OP', and the 



locus of P is a straight line. 

If in the hypocycloidal straight line motion we consider two 
diametrically opposite points of the small circle, we see that they 
will describe straight lines at right angles to one another. There- 
fore if we guide the centre of the small circle by means of a link 
CO (Fig. 97) and guide R through a short distance RS by means 
of a sliding pair, the motion of all points of circle C will be deter- 
mined, and P will move in a straight line AB through quite a long 
distance for a very small motion of A 3 . This also belongs to Class 2. 



144 



MACHINERY OF TRANSMISSION 



The angle a in Fig. 98 should not exceed 20 , otherwise the 
sliding at R becomes excessive. If a = 20 , s = -| of 2a = i\a 
(nearly). During such a stroke R travels through a distance 
p = 2<z(i — cos a) = 2a x .06 = .1 2a = .09^. The use of the sliding 
pair at R is sometimes avoided by carrying this point on a vibrat- 
ing pillar (Fig. 99). The motion now comes under Class 3. If the 
length of the pillar is equal to the stroke, and the angle a is equal 




Fig. 96 



to 20 , the maximum deviation from a straight line will be only 



4T00" °f tne str °ke. 



It was seen that if the middle point of the link PR be chosen 
as the point of attachment of the link CO, the path of C was a 
circle. If any other point be chosen, its path will be an ellipse. 
The elliptic arc can then be approximated by the arc of a circle, 
and an approximate parallel motion is thus obtained. Fig. 100 
shows the paths of the various points of the line PR, and also the 



PARALLEL MOTIONS 



145 



radii and centres of the approximating circles. This form belongs 
to Class 4. If R is carried on the end of a vibrating pillar, it 




belongs to Class 3. The constrainment of P to move in a straight 
line may be accomplished in other ways. Two points may be 




Fig. 98 



chosen in the plane of the small circle, and the elliptic arcs 
approximated by circles. This is Robert's parallel motion. 



146 



MACHINERY OF TRANSMISSION 




PARALLEL MOTIONS 



147 



C. The Conchoidal Straight Line Motions 

If a line moves, so that one of its points lies on a straight line, 
while the line itself always passes through a fixed point, all points 
of the line describe curves known as Conchoids. We can obtain 
the equation of the curve as follows : let AB (Fig. 101) be the 




Fig. ioi 



straight line, and G the fixed point. From G drop a perpendicu- 
lar GO on AB, and take O as the origin. Let P X G be the 
moving line, and consider the conchoid as traced by point C 
whose coordinates are x and y. Then 



148 



MACHINERY OF TRANSMISSION 



y = (b — x) tan«, 



tan a = 



V* 2 



V^-x 2 



y= W-x) — - — » 

xy = (b — x) V* 2 — xr- 

If e is positive, and less than b, the curve is such as is described 
by C. If e is positive and equal to b, we have a cusp at G. If e 
is positive and greater than b, the curve is described by a point 
C 2 . If e is negative, C 3 describes the curve. The curve can be 
used as a straight line motion by constraining points such as C lf 
C 2 , or C s> to move in the arc of the conchoid, and also by con- 




Fig. 102 

straining the line to pass through G. The conchoidal arc must be 
approximated by means of the arc of a circle. The point C 2 is 
the best one to constrain, as the looped portion of the curve is very 
nearly a circle where it cuts the axis of X. (See Fig. 102.) At 

that point the radius of curvature is equal to p=- -■ All 

e 
conchoidal guides belong to Class 4. 



PARALLEL MOTIONS 



149 



D. The Lemniscate Straight Line Motions 

If two points C and D of a line are made to describe circles, 
then any other point of the line will describe a peculiar looped 
curve known as a Lemniscoid. This curve is of the fourth degree, 
and a special form known as the Lemniscate, where the radius bars 
are equal and the tracing point is midway between C and D, is 
shown in Fig. 103. CD is the moving line, and A and B are the 




Fig. 103 



fixed centres about which C and D revolve. P is the tracing point. 
That portion of the curve which is traced when the arms are 
parallel is of peculiar interest. Examination shows that for quite 
a distance on either side of this position the curve approximates a 
straight line. At the instant the arms are parallel, the line CD is 
turning about a centre at an infinite distance, and is therefore 



150 MACHINERY OF TRANSMISSION 

moving parallel to itself. We may consider the curve near the 
double point O as a straight line, and attach a cross-head or other 
portion of a machine to it. The conditions usually chosen are 
that the guided point should not only pass through the middle but 
also through the extreme positions of the rectilinear path. James 
Watt used this parallel motion to guide the cross-head of his steam- 
engine. He did not attach the cross-head directly to P, however, 
but duplicated the motion of P by means of an ordinary panto- 
graph. All lemniscate straight line motions belong to Class 3. 

E. Inversors 

The first straight line motion discovered which fulfilled the con- 
ditions of Class 1 was based on the properties of inverse curves, 
particularly those of the circle. If P (Fig. 104) be a fixed point, 
UY any curve, and PB a radius vector intersecting UY dX A, then 
the curve WX so constructed that PA x PB = PA' x PB' = 




Fig. 104 



constant, is called the inverse of UY with respect to /'as a pole of 
inversion. It is shown by modern geometry that if UY is a circle, 
the curve WX is a circle also, and if the circle UY passes through 
P, WX becomes a straight line. This last proposition can be 
shown very simply. Let (P Fig. 105) be the pole of inversion. 
The circle A'AP, which we wish to invert, passes through it. 



PARALLEL MOTIONS 



151 



Draw BB' perpendicular to the diameter through P. Then by 
similarity of triangles, 

PA:PB'::PA':PB, 

or PA x PB = PA' x PB' = constant. 

Hence BB' is the inverse of the circle through P. 




Fig. 105 



Peaucellier's Straight Line Motion. — This consists of two equal 
links PD and PE hinged at a fixed point P. The extremities D 
and E are connected together by a rhombus of hinged links 
AD BE (Fig. 106). The lines PAB and DE intersect in the 
centre of the rhombus at C. We have 

PA = PC- CA = /cos a - m cos /?, 

PB = PC+ CA = /cos a + m cos (3, 

PA x PB= P cos 2 a - m 2 cos 2 (3. 

But also /sin a = ;;/ sin /?, 

pi 2 cos 2 a = ;;r — / 2 sin 2 a. 



152 



MACHINERY OF TRANSMISSION 



Hence, PA X PB = / 2 cos 2 a — m 2 + / 2 sin 2 a = P — ?n 2 = constant. 
Therefore if we move ^ on any curve, B will move on the inverse 
curve with respect to P as a pole of inversion. Now if we con- 
strain A to move in the arc of a circle by means of a link PA, B 
will move in a circle also. If PA is less than PP, B will travel in 
a circle whose concavity faces away from P. If PA is greater 




Fig. 106 

than PP, B will travel in a circle whose concavity faces toward P. 
If ^4i? equals PP, so that the circle in which A moves passes 
through P, B will describe a straight line. It is easy to find an 
expression for the radius of the inverse circle in terms of known 
constants. Call the constant product PA x PB p 2 . Then in Fig. 
107, 

PA xPB = PA' x PB' = p 2 , 



PB = -£- 
PA 



PB' 



PA 



PB'-PB^2r'=-£- — , 

PA' PA' 



PARALLEL MOTIONS 



153 



where r* is the radius of the inverse circle. If we denote by r the 
radius of the original circle, and the distance of its centre from P 
by a, we have 



2r 



, _ p 2 {PA - PA') 
PA X PA' 



2rp- 



PA x PA' 




Fig. 107 



Hart's Straight Line Motion. — Let the four rods GH, FK, 
GF, and KH (Fig. 108) be jointed together as shown. The figure 
FGSKH is called a complete parallelogram when FK — GH = L, 
and GF= KH = I, which two lengths become constants of the 
chain. Call the distance GK = x and FH = y. Then, as the 
form of the apparatus is changed, x and y will vary in length. 
Since the triangles GFH and KFH are equal (all three sides 
being equal), angle KHF is equal to angle GFH, which may be 
called /?. Also the lines x and y are always parallel, since / sin (3 
expresses their distance apart at either end. Take any fixed point 
in GF such as O, and through it draw a line parallel to x and y. 
Call OF=a. The line will evidently cut KH at R at a fixed 
distance a from H It will also cut the other two links at P and 

Then b = a — and b is constant, or P is a 



Q. Qz\\ FF=b, 



154 



MACHINERY OF TRANSMISSION 



fixed point on link KF. Similarly Q is a fixed point on GB, and 
QB = 



Now OP=x-, and OQ=y- — -; hence, 



But x = L cos a — I cos ft, 

y = L cos a + / cos ft, 
xy = D cos 2 a — P cos 2 ft = L 2 cos 2 a — I 2 (/— sin 2 ft). 



G^ 







From the figure 
or 



/sin a = L sin /?, 

. L . 

sin p = ^ sin a, 



xy = Z 2 cos 2 a - / 2 (7- ^ sin 2 ( 1\ = Z 2 -I 2 . 
Hence, OPxOQ = (L 2 - P) a AL^L = constant. 

Or, if the apparatus be hinged at O, P and Q will describe 
mutually inverse curves with respect to O as a pole of inversion. 
If P is made to move in a circle whose circumference passes 
through O, then will Q describe a straight line. 



CHAPTER VI 

TRANSMISSION OF MOTION BY CONTACT WHEN DIRECTIONAL 
RELATIONS ARE NOT CONSTANT 

i. CAMS 

A cam is a rotating body, which transmits motion to a follower 
by means of a curved edge. Usually the conditions do not 
involve any variable velocity ratio, but a series of positions of both 
driver and follower being given, as well as the outline of one, the 
form of the other is constructed. 



A. Disk Cams 

In the disk cam the motions of the driver and follower take 
place in the same plane. There are two principal kinds of disk 
cams. One transmits motion to 
a pin or roller, and the other to 
a straight bar. In the first case 
(Fig. 109), let O be the centre 
about which the cam revolves, 
and let OQ be the path of the 
pin or roller. A, 1, 2, 3, and 
4 are given positions of the 
centre of the pin corresponding 
to positions OA, Oi', O2', O 
3', and O 4' of the cam radius. 
About O as a centre, and with 
a radius Oi, draw an arc OB 
intersecting O 1' in B. Then B 
is a point on the pitch curve 
of the cam. In like manner 
points C, Z>, and E are ob- Fig. 109 

!55 




56 



MACHINERY OF TRANSMISSION 



tained. It is readily seen that the point A can be brought to 
rest at any time by giving the cam a circular outline about O as 
a centre. Having thus found the pitch curve, the working edge 
is obtained by drawing a parallel curve at a constant normal dis- 
tance from it equal to the radius of the roller. If A is started 
from rest, care should be taken not to give it too great velocity at 
first ; in other words it should be given, if possible, a gradually 
increased velocity, and the same care should be observed in 
bringing A to rest. If A is given a constant acceleration during 
the first part of its travel, the equation of the cam curve can be 
worked out easily if the angular velocity of the cam is constant. 

The constant distance 
OA we may call a. 
Then 

Ai=\ct\ 
where c is the constant 
acceleration. The an- 
gle BOA = 6, hence 

^ - = q) = constant, 

r= OB = a + ±ct 2 



or the polar equation 
of the curve is of the 
form 

In a similar method the equation of that portion giving a constant 
retardation can be worked out. Care should be taken in this case 
that c is not greater than g, the gravitation constant, otherwise the 
follower will leave the cam curve unless some other force than 
gravitation is employed to cause its return. 

If QA produced does not pass through O, we proceed as follows : 
let A, i, 2, 3, and 4 (Fig. no) be positions of the follower corre- 
sponding to positions OA, Oi , ,02 l } 0^', and O4' of the cam. Draw 




Fig. 1 10 



CAMS 



157 



OP through A. Construct arcs about O as a centre cutting the 
radii in B, C, D, and E, and the line OP in W, X, Y, and Z 
On arc 1 B lay off BR equal to IV 1, on arc 2 C lay off CS equal 
to X2, etc. Then obviously ARSTU will be the pitch curve of 
the cam. 

In the second case of disk cam, the motion which can be given 
to a follower is not so universal as that which can be given by the 
first. The manner of constructing the cam curve to satisfy a given 




Fig. hi 



condition is as follows: let AA", n", 22", 33", and 44" (Fig. 
in) be successive positions of a bar which moves parallel to itself 
in the direction OQ. Let the corresponding radii of the cam which 
successively occupy the position OQ be OA, Oi', O2', O3', and 
O4'. Suppose we bring the cam to rest by giving the whole an 
equal and opposite rotation about O. The lines n", 22", 33", 
and 44" will take up the positions XX", YY", ZZ", and UU", 
where these are drawn at right angles to the several radii, from 



158 , MACHINERY OF TRANSMISSION 

points projected by circular arcs from 1, 2, etc. But these lines 
must all touch the cam, hence they form the envelope of the 
required curve. That these conditions cannot always be fulfilled, 
can best be seen by referring to Fig. 112. If the line which forms 
the envelope intersects its previous positions in points 1, 2, 3, 4, 




Fig. 112 

etc., one beyond the other, the cam curve can be cut out of metal. 
But if any intersection falls between two previous ones, the con- 
struction is impossible, as a cusp is formed at the point where the 
first two intersections coincide. 

If the straight bar rotates about a fixed centre instead of moving 
with pure translation, the same general method would be pursued. 



B. Cylindrical Cams 

Here the motion of the follower is in a plane at right angles to 
that of the driver. A helical slot or groove of the desired form 
is cut in the surface of a cylinder, and is made to engage a pin 
or roller whose diameter is equal to the width of the slot. The 
cylindrical surface of the cam must here be developed on a plane 
and the motion of the follower be considered as referred to it. In 



CAMS 



159 



Fig. 113 is shown the complete solution of a case of the cylindrical 
cam. The developed projection gives the best idea of the general 




form of the slot, but the curve can be reprojected on the original 
cylinder as shown if required. 



PART III 

MECHANICS OF THE STEAM-ENGINE 



CHAPTER I 

KINEMATICS 

i. GENERAL DESCRIPTION OF THE STEAM-ENGINE CHAIN 

In its most elementary form the steam-engine consists of six 
links as shown in Fig. 114. The bed or frame is represented by a, 
and to this the motion of all other links is referred ; b is the crank, 
including the shaft, eccentric, and fly-wheel ; c is the piston, piston- 
rod, and cross-head ; d the connecting rod ; e the eccentric rod ; 

Q 

I 




Fig. 114 



and f the valve. This compound chain may be divided into two 
simple ones, the crank and engine bed being common to both, and 
the whole reduces to two chains like Fig. 5. It is the purpose of 
this chapter to discuss the relative motions of the various parts of 
the steam-engine chain, and to find the general proportions and 
positions of the parts which will give the best results. 



2. THE PISTON-CRANK CHAIN 

This chain consists of four links, — the engine bed, the crank, the 
connecting rod, and the piston. Referring all motions to the engine 
bed, the motion of the crank is one of pure rotation, that of the 
piston is pure rectilinear translation, while that of the connecting 
rod is a combination of both. 

163 



164 



MECHANICS OF THE STEAM-ENGINE 



A. Relation between the Position of the Crank and the 
Positions of Other Points of the Chain 

The position of a point in the connecting rod for any given 
angular position of the crank can be expressed as follows : Let 
O (Fig. 115) be the centre of the crank shaft. Since the motion 
of the piston is identical with that of the centre of the wrist pin, 
we may consider the whole of the piston and cross-head as con- 
centrated there. Consider the position of any point R of the 




Fig. 115 

connecting rod. Call its coordinates referred to an origin at O 
and axis of X in the line of connection, x and y, and call its fixed 
polar coordinates referred to pole at P and vectorial angle 
measured from PQ, r and 8. 6 is the variable crank angle, a the 
corresponding angle of the connecting rod, L the length of the 
connecting rod, and / that of the crank. Then 

x — I cos 9 — L cos u + r cos (a — 8) , 

y= rsin(« — 8). 

/ sin = L sin a, sin a = — sin = n sin 6, 



Since 



cos a = — V 1 — n 2 sin 2 0. 



Expanding the expression for x, 

x = I cos 6 — L cos a-\-r cos a cos 8 + r sin a sin 8. 



THE PISTON-CRANK CHAIN 165 



And substituting for a in terms of 6, 
I — nr cos 8 



x = I cos 6 + ( Vi — n 2 sin 2 6 -f rn sin 8 sin 0. 



In the same way, 

y = r sin a cos 8 — r cos « sin 8 



= 772 cos 8 sin 8 + r sin 8 Vi — ^ 2 sin 2 0. 

These are the most general expressions for the coordinates of a 
point of the connecting rod in terms of the variable angle 0. When 
the point R lies in the axis of the rod, 8 = 0, and the equation 
reduces to 

x == / cos + (/ ~ nr) Vi - n 2 sin 2 0, 

n 
y = rn sin 6. 

Finally, if r — L, we get the position of the crank pin referred to 
O as an origin, 

Xq = / cos 6, 

jy Q = /sin 0. 

The coordinates of other points of the crank can be similarly ex- 
pressed in terms of their angular position and distance from the 
centre of rotation. 

Now if r= o, the position of Pis defined, or 



x P == / cos + - V 1 - n 2 sin 2 6, 
y P = o. 

If the stroke of the piston is supposed to take place along the 
diameter C'H' of the crank orbit (Fig. 116), its position 
measured from the middle of its stroke will be 

= /cos0 + -(Vi-/* 2 sin 2 0-i). 

These last, which are the most important of all, express the posi- 
tion of the centre of the wrist pin as a distance measured from the 



1 66 



MECHANICS OF THE STEAM-ENGINE 



centre of the shaft or from the middle of the stroke. But piston 
positions are often measured from one or the other ends of the 
stroke, or from H and C (Fig. 116). Calling these distances in 
general z, 



= /( i - cos 0) + -(i - V i - n 2 sin 2 $), 



/(i + cos ^)--(i-Vi-« 2 sin 2 0). 




Fig. ii6 



If finally the connecting rod be of infinite length, n = o, the 
motion of P becomes simply harmonic, and substituting n — o in 
the expression for z, and evaluating the indeterminate form, we 

have 

X P = 00, Xp = / cos 0, 

z H =l{\ —cosO), 

Z c =/(l + COS0). 

It is often useful to lay off x F ' as a polar curve, using OR as a 
radius vector. Calling p = x F ', the equation of the curve is 



P = /cos<9-f--f V i — n 2 sin 2 6— i), 

and its form is shown in Fig. 117. If n = o, the equation reduces 

to the simple form 

p = I cos 0, 



THE PISTON-CRANK CHAIN 



167 



which is the equation of a circle whose diameter is /, whose centre 
lies on the axis of X to the right of the origin, and whose circum- 
ference passes through O. 

The graphical solution of all the preceding cases consists in 
nothing more than the construction of an accurate drawing of the 




Fig. 117 

chain for the chosen value of 6, from which x or z can be scaled 
off. If the stroke of the piston be considered to take place on 
C'H' (Fig. 116), the position of the piston can be easily found 
at E' by projecting Q on a circular arc from a centre on the line 
of connection and with a radius equal to L. Similarly, if the 
length of the rod be infinite, the position will be at E> found by 
straight projection from Q. 

B. Relation between the Position of the Crank and the 
Velocities of Other Points in the Chain 

The component velocities of R, parallel to the axes of X and Y, 

can be found by evaluating — and — -, after which the resultant 

velocity can be found both in magnitude and direction. We have, 



/t—nr cos $\ 
x — l cos + [ - )Vi-» siir V + rn sin sin 6, 



from which 

dx , . a d0 n sin 6 cos jq . m 

= —/sm6——n(/—nrcosd) —7 ===== p rn sin 8 cos 6 — . 

dt dt v Wi-n 2 sin 2 6dt dt 



1 68 MECHANICS OF THE STEAM-ENGINE 



And also y = rn cos & sin + r sin 8 Vi — ri 2 sin 2 0, 

dy . A d0 „ . « sin cos ^9 

— = ;tz cos o cos 6 rnr sm o — — — 

<# <# Vi -n 2 sin 2 6 dt 

In nearly all cases of steam-engine analysis, the fly-wheel is suf- 
ficiently large and heavy to make — practically constant. Calling 

dt 

this w, the formulae reduce to 

dx 7 • a // s\ sin cos , . « A 

— = — /o) sm — fna{l— nr cos o)- + rtna sm 8 cos 6, 

dt -Vi — n 2 sin 2 

-=£ = r;z<D cos o cos — *72 J o> sin o — — 

dt Vi-7* 2 sin 2 

These are the general expressions in terms of 6 for the component 
velocities of any point in the connecting rod, whose position is 
defined as before. The resultant velocity will be 



*=V(fM;)' 



and the angle which its direction will make with the axis of X 
will be 

dx 

Now if the point R lies in the axis of the rod, S = o, and the 
velocities are 

dx j • a / / \ sin cos 

lui sin — n\l— nr) m 



dt Vi - n 2 sin 2 6 

^ = rno>cosO. 
dt 

If r = Z, we get the component velocities of the crank pin, 

u x = — /(o sin 6, 

and u y = loi cos 0. 






THE PISTON-CRANK CHAIN 



169 



The component velocities of all other points in the crank can be 

similarly expressed in terms of their angular position and distances 

from the centre. 

Finally if r = o, we have the velocity of P, or that of the centre 

of the wrist pin, 

7 ■ n 7 sm cos $ 
v x = — Ao sm — n/o> — » 

Vi — n 2 sin 2 



And the velocities of all other points of the piston and cross-head 
will be the same. 

This most important velocity can be expressed as the ordinate 
of a curve in rectangular coordinates, whose abscissae are piston 




Fig. 118 



positions. Taking a number of arbitrary values of 6, the corre- 
sponding values of x and v x can be calculated, and the results 
plotted in graphical form along the stroke of the piston or the 
diameter of the crank orbit. (See Fig. 118.) 

When the length of the connecting rod is infinite, compared 
with the throw of the crank, n becomes equal to zero. Substi- 
tuting this value, 

v x = — /o) sin 0. 

In this case can be simply eliminated between the equa- 
tions of position and velocity. Referred to the diameter of 



170 



MECHANICS OF THE STEAM-ENGINE 



the crank orbit as a stroke, the position of the piston measured 
from O is 

Xp = I cos 0. 

Eliminating 6, 




This is the equation 
of an ellipse referred 
to its principal axes, 
as is evidently the 
necessary result of a 
simple harmonic mo- 
tion. The angle is 
the eccentric angle 
(Fig. 119). 
The piston velocity, viz., v xi may be laid off along the crank OR 

in such a way as to form a polar curve of piston velocities. Calling 

the radius vector o-, the equation of the curve is 

, . n 7 sin $ cos 6 
a- = — /(o sin — n/u) — . 

This curve is shown in Fig. 120. 

When n =0, 



Fig. 119 



V 1 — n 2, sin 2 
o- = — /a) sin 9, 




Fig. 120 



the equation of a circle whose centre lies on the axis of Y, below 
the origin, whose diameter is /, and whose circumference passes 
through the origin. 



THE PISTON-CRANK CHAIN 



171 



The angular velocity of the crank is, as we have said, constant 
and equal to w. 

The angular velocity of the connecting rod is found by evaluat- 
ing — . Since a = sin _1 f n sin 0\, 



dt 
we have 



da _ — n<D cos 9 
dt Vi-^ 2 sin 2 



Graphical constructions for the above velocities are, in general, 
simpler and more useful than the analytic. Produce the axis of 




Fig. 121 



the connecting rod to cut the axis of Y at S (Fig. 121). Let /be 
the instantaneous centre of the connecting rod referred to the bed 
of the engine. Then 

velQ _u _IQ 



velP 



But from similarity of triangles — — = — ^ ; 



hence, 



v x = u 



OS 

0$ 



and since u (=/w) and OQ (=/) are constants, v x \s propor- 
tional to OS. Furthermore if we choose our scale of velocities so 



172 



MECHANICS OF THE STEAM-ENGINE 



that u is laid off equal to the crank throw, OR becomes equal to u, 
and OS = v x . 

Another simple construction is to lay off u from Q along QI. 
Through its extremity draw a line parallel to the connecting rod, 

and where this cuts IP will give v x , since — = *=-. In either case 

v x PI 
the curve of velocities can be simply drawn in. 

When the length of the connecting rod is infinite, OS is simply 

the projection of OQ on the axis of Y, and the velocity curve 

becomes a circle when the scale is taken such that u = OQ = /. 




Fig. 122 



This is seen also from the analytic expression, for, when Ao = / by 
reason of choice of scales, the equation of the ellipse reduces to 
that of a circle.* 

The angular velocity of the connecting rod can also be graphi- 
cally constructed by employing the instantaneous centre method. 
By reference to Fig. 16, Part I, the following construction can be 
verified. Lay off from / in any direction the constant angular 



* Care should be taken here to avoid confusion. It is not intended that 
since fe = /, a, = I 5 but that « - * - i scale of len ^ hs 



1 x 



scale of velocities 



THE PISTON-CRANK CHAIN 1 73 

velocity of the crank. From the extremity of this line draw a line 
through Q. From O draw a line parallel to w to meet the one 

last mentioned, and this will give the magnitude of — . Its direc- 

dt 

tion can be told by simple inspection of the drawing. This con- 
struction is shown in Fig. 122. 

Another simple solution, where the velocities of both P and Q 
are given, is also shown on the same figure. Bring P to rest by 
giving the whole system a velocity equal and opposite to v x . Com- 
bining — v x with u at Q, we get QN, the absolute velocity of Q 

referred to the piston. This is L — , from which — can be obtained 

dt dt 

by simple division, 

C. Relation between the Position of the Crank and the 
Accelerations of Other Points in the Chain 

The component accelerations of R parallel to X and Y can be 

d x d v 

found by evaluating —=-$ and — -^ We have, where o> = constant, 

dx 7 • a / 7 *\ sin cos . <> „ 

. — = — /o> sin — «a)(/— nr cos 0) — — -+- rina sin 8 cos 0, 

dt Vi-»*sin 2 

d 2 x _ 

72 a 2/7 &\ f cos 2 0+n 2 sin* } a - * • /1 

— /V cos 9—n<a\I— nr cos 6) 4 ! I — rW sin 8 sin 0. 



And also l (i-« 2 sirftf) 

//y a /i 2 • & sin 9 cos 6 

->l — moi cos 6 cos V — r;ra> sin — — , 

dt Vi-/z 2 sin 2 <9 

d 2 y 2 a • a 2 2 • s f cos 2 + ?z 2 sin 4 1 
— j£ = — rntf cos 8 sin — rn'm* sin ^ ! }. • 

dt I (i-?z 2 sin 2 0)* J 

These are the general expressions in terms of 0, for the com- 
ponent accelerations of any point of the connecting rod whose 
position is defined by r and 8 as before described. The resultant 
acceleration will be, 



174 



MECHANICS OF THE STEAM-ENGINE 



in magnitude, and its direction will be defined by the angle e, 

where 

d 2 y 

,~d£ 

£==tan 35" 

dt 2 

If the point R lies in the axis of the rod, 8=0, and we get, 

d 2 x jo a . 2/7 \ r cos 2 + n 2 sin 4 

— -9 = — /a) J cos 6 + noi\l— nr) ■ 



(i-« 2 sin 2 6*)2 J 



^=-,Wsin0. 
dt 



If r = Z, we get the component accelerations of the crank pin, 
q x — — /(o 2 cos 0, 
and <7„ = — /w 2 sin 0. 




Fig. 123 



The accelerations of all other points of the crank can be similarly 
expressed in terms of their angular positions and distances from 
the axis of rotation. 

Finally, if r = o, we have the acceleration of P or of the centre 
of the wrist pin, 

7 2 a 2/ f cos 2 + n 2 sin 4 \ 
p x = — loi 2 cos 6 — nw 2 / \ — \ , 

I (i-?* 2 sin 2 0p J 

and p y = o. 






THE PISTON-CRANK CHAIN 



175 



and all other points of the piston and cross-head have the same 
acceleration. All these accelerations are of the utmost importance 
in connection with the dynamics of the steam-engine. 

By choosing arbitrary values of 0, and computing the corre- 
sponding values of p x and x' p , a. curve can be drawn with reference 
to rectangular coordinates, which shows by its ordinates the magni- 
tude of the piston acceleration and by its abscissae piston positions. 
Such a curve is shown in Fig. 123. When the length of the con- 
necting rod is infinite compared with that of the crank, n=o, 
and the motion becomes simply harmonic, as in that case, 

p x = — /o> 2 cos 0. 

Here we can easily eliminate between this, and the equation for 
piston position, which is, 

x' p = /cosO, 
and we have 

, /(x) 2 X o 

A= — = -xo> 2 , 

the equation of a straight line through the origin. 




Fig. 124 

We may also lay off /„ as a radius vector along the crank throw, 
and thus construct a polar curve of accelerations. Calling the 
radius vector t, the equation of the polar curve would be, 

72 a 27 f cos 2 + « 2 sin 4 1 

T = — f(x) COS — «0)V \ [ . 

I (i-n 2 sm 2 $y ) 
This curve is shown in Fig. 1 24. 



176 



MECHANICS OF THE STEAM-ENGINE 



If n = 0, the equation reduces to 

T = — /a) 2 cos 6, 

the equation of a circle whose diameter is /, whose centre lies on 
the axis of X to the left of the origin, and whose circumference 
passes through the origin. 

The angular acceleration of the connecting rod can be found by 
simply differentiating the expression for its angular velocity, 

da _ — nay cos 
~dt~ 



Vi 



sin 2 



d 2 a n(i — n 2 )u) 2 sin 6 

df 1 (i-n 2 sin 2 6)% ' 

o) being taken as constant. 

Graphic solutions of the general case can best be done by com- 
bining the solutions of the special cases of the two ends. The 

graphic solution of 
the acceleration of 
the crank pin Q 
follows immediately 
from the analytic 
equations for q x and 
q y . For if we lay 
off a distance from 
Q toward O equal 
to-/o> 2 (Fig. 125), 
the projections of 
this length on the 
coordinate axes will 
give the component 
accelerations of Q as 
these are evidently 
If the scale of accelerations is so 




Fig. 125 
I J- cos 0, and — /o> 2 sin 



chosen that — /w 2 is equal to OQ, the component accelerations 
will be merely the projections of the crank -throw itself upon the 
coordinate axes. 



THE PISTON-CRANK CHAIN 



1 77 



The graphic solution of the acceleration of the wrist pin is more 
complex. We must here again employ the instantaneous centre 
of the connecting rod with respect to the bed of the engine. 
Since OS (Fig. 121) is proportional to the velocity of P, and since 
OS is measured from a fixed point O and in a fixed direction 
O Y, the velocity of S along O Y will be proportional to the accel- 
eration of P. Imagine the line Qw (Fig. 126) as moving with 







.J 1 




"" '/' ' 




*-"^ /' ' 








,"'' '' / 








''' '' / 








,-' " ' 








^'*' s' S ! 








»•'■'"' s* ' 


u 


'■**' y' S 


tf-A 


K ''' /' 9 




\ T> 


"^ >X / 


u^>~^ 


-^\ / ; 


/^ s 


^^^>fcl 




A \ ^^^^-J** 







\h 



Fig. 126 



PQ, being in fact a prolongation of the axis of the rod. Then S, 
when considered a point of PQ, is moving in a direction SJV per- 
pendicular to SI, the instantaneous radius. If we project Q to M 
on the arc of a circle about / as a centre, the velocity of M will 
evidently be equal to that of Q, and at right angle to SI. This 
velocity u can then be laid off, and from it the magnitude of the 
velocity of S follows by obvious construction. Now if this velocity 



i/8 



MECHANICS OF THE STEAM-ENGINE 



SJV of S be resolved along the axis of Y and along PQ, the 
former will evidently be proportional to the acceleration of P. 
This proportionality becomes an equality when as before the 
scale of accelerations is so chosen that the acceleration of Q, viz., 
— /o> 2 , is taken equal to OQ. 

This choosing of the proper scales of velocity and acceleration 
sometimes give rise to confusion in the mind of the student. All 
such trouble can be avoided by the following simple device. The 
analytic expression for the acceleration of P is 



p x ~— /o> 2 cos 6— noi 2 l \ 



{ cos 2 4- n 2 sin 4 1 



(i -« 2 sin 2 0)^ 



At the head-end dead point, i.e. when = o°, this reduces to the 
simple form 



and at the* crank-end dead point to 

A]^180° = + ^ 2 (l 



n), 



which expressions can be simply calculated Now in the preced- 
ing construction when the crank pin Q arrives at one of the dead 




Fig. 127 

points, H, for instance, / coincides with P, M and Q with B, and 
S and w with O. The construction in this case is as in Fig. 127, 
where Hu is the velocity of the crank pin, and ON H the accelera- 



THE PISTON-CRANK CHAIN 



179 



tion of the cross-head. Then if we calculate this acceleration 
from the simplified analytic expression, lay it off to any convenient 
scale as ON H , and draw N H P H , the proper scale of velocities 
becomes immediately known as Hu, and this may now be used all 
around the crank orbit to get other values of the acceleration of P 
on the same scale as ON B . 

Having now the accelerations of both ends of PQ, it is easy to 
determine that of any other point R (Fig. 128). Project R on the 
arc of a circle about P as a centre to A, and consider first the 
acceleration of A. The absolute acceleration of Q is QT equal to 
— /to 2 directed inward along the radius vector. Bring P to zero 
acceleration by giving the whole system an acceleration equal and 




Fig. 128 



opposite to that of P. Combining — p x with — lay 2 at Q, we get 
Qw, the resultant acceleration of Q referred to origin P in the 
piston. The component of Qiv at right angles to QP, viz., Qb, is 
evidently a tangential component of acceleration about P, and 

d 2 a 
Qb= L — -, from which the angular acceleration of the connect- 
dt- 

ing rod can be found by simple division. The component of Qw 



along the rod, or Qa, is a normal component, or Qa = — L 



del 
df z 



from which the angular velocity might be determined. Now the 
tangential and normal components of all points of the body being 
simply proportional to their respective distances from P, when P 
has no acceleration, it follows that the components of A are equal 



180 MECHANICS OF THE STEAM-ENGINE 

to those at Q when multiplied by—- Hence the construction as 

given in Fig. 128. The components at R will be equal to those 
at A, but their directions are of course different, being along and 
at right angles to RP. The resultant Rw' is the absolute accel- 
eration of R referred to the piston and origin at P, and combining 
this with +A> we get RS, the absolute acceleration of R when 
referred to the bed of the engine both in magnitude and direction. 
The projection of this on lines parallel to the coordinate axes gives 

^and^. 
dt 2 dfi 

3. VALVE GEARING 

A. Relation between the Position of the Crank and 
the Position of the Valve 

The chain of links forming the valve gearing consists, as is seen 
in Fig. 114, of four links, which are the engine bed, the eccentric, 
the eccentric rod, and the valve. If we express positions, veloci- 
ties, and accelerations of points of this chain in terms of an angu- 
lar position 61 of the eccentric, the results will be identical with 
what has preceded, and if expressed in terms of an angular posi- 
tion 6 of the crank, a phase angle, viz., angle QOC (Fig. 114), 
must be introduced. 

The valve of a steam-engine performs the duty of admitting 
steam alternately into the ends of the steam cylinder, and of 
exhausting the same into the atmosphere or the condenser. In 
the design of the valve gear the problem is to open and close the 
ports at certain given positions of the piston. It is therefore 
the position of the valve with which we are principally con- 
cerned, its velocity and acceleration being of minor importance 
at present. 

{a) The Plain Slide Valve 

The motion of the slide valve is usually produced by means of 
an eccentric, which consists of a circular plate, keyed to the shaft 
in a plane at right angles to its axis, and surrounded by a strap 
which is fastened through the eccentric rod to the valve. It is of 



VALVE GEARING 



181 



course evident that the eccentric acts exactly as a crank, whose 
throw is equal to the distance between the centre of the shaft and 
that of the eccentric 
sheave. In Fig. 129 
OC is the throw of 
the eccentric, and 
it is always denoted 
by r. The valve 
which it operates 
will move through 
a total distance 2r 
when directly con- 
nected. 

In Fig. 130 let 
OQ be the crank, 
and OC the eccen- 
tric of an engine whose shaft is at O. The invariable angle QOC 
between the two we shall call /?. If 6 X is the angle between the 
line of connection of the valve and the throw of the eccentric, 
then the position of B can be written 




Fig. 129 



xj = r cos 0! + - { Vi — n 2 sin s 



C? - 



— I 




i-*i-H 



Fig. 130 



when the distance x B ' is measured from the middle of the stroke 
of B. Now if 1 = 4- ft, we have, on substituting this, 



x B ' = rcos(0 + P) + -\ Vi-« 2 sin 2 (0-r-/3) 



Fig. 131 is a diagrammatic representation of a plane slide valve 
in its central position with respect to its travel, i.e. when B is at 
O' (Fig. 130). The steam ports 5S" are separated from the exhaust 



182 



MECHANICS OF THE STEAM-ENGINE 



port E by means of the bridges bb. The valve is always made 
longer than the sum of the widths of the three ports and two 
bridges. The amount by which the outside edges of the valve 
project beyond the outside edges of a steam port when the 
valve is in its central position of travel is called the Outside Lap, 




Fig. 131 



and is always denoted by e. The amount by which the inside 
edges of the valve project beyond the inside edges of a steam port 
when the valve is in the same position, is called the Inside Lap, 
and is always denoted by i. These laps are given to the valve to 
work the steam expansively. They may or may not be the same 
at the two ends. 

Fig. 132 shows the relative positions of the crank, eccentric, 
and valve for the head-end dead point. As the valve moves very 




Fig. 132 



nearly symmetrically across the face of the ports, in order that the 
steam distribution may be nearly alike on the two sides of the 
piston, the middle position of the valve referred to its own travel 
will coincide very nearly with the middle line AA of the ports. 
The middle position of the valve will evidently come when the 



VALVE GEARING 1 83 

eccentric throw r makes an angle of about 90 with the direction 
of motion of the valve, which is usually identical with that of the 
piston. Hence if the valve is made longer than the sum of the 
three ports and two bridges, the eccentric must be drawn past 
this position when the crank is on the dead point, as at that time 
the steam port must be just opening on the head end as shown. 
In most engines, especially if high speed, the valve is made to 
open a little while before the crank arrives at the dead point. 
The small amount by which the steam port is open when the 
crank turns the centre is called the Lead, and is denoted by v. 
If we denote by 8 the angle between the eccentric and the per- 
pendicular to the line of valve connection, then, if the piston and 

valve move parallel, 

P = 9 o° + 3, 

and 8, a constant of the valve gear, is called the Angular Advance, 
and its exact value will be determined hereafter. 

Substituting this value for /? in the equation for valve position, 
we get 

x B ' = -r sin (0 + 8) + - (Vi - ;rcos 2 (<9 + 8) _ 1 ) . 

Now in a valve design the second term of the equation can be 
generally neglected, as will be seen from the following case. 

Let r — 2, and n — .04 (ordinary dimensions). Then the maxi- 

mum value of the first term will occur when (0 + 8) = -, and then 



7T 
2' 

r sin (0 + 8) = — 2. The maximum value of the second term 



will occur when cos 2 (0 -f- 8) is equal to unity, or 



- ( Vi - n 2 cos 2 (0 + 8) - 1) = — ( Vi - .0016 - 1) = - .04, 
n v v J .04 x 

which is small enough compared with 2.00 to be neglected for 
purposes of design. Hence we shall always write 

x B ' = — rsm(8 + 6), 
a simple harmonic equation. 

Expanding the right hand side, we get 

xj = — r cos 8 sin 6 — r sin 8 cos 0. 



1 84 



MECHANICS OF THE STEAM-ENGINE 



But r sin 8 is a constant, as is r cos 8 also. Calling these A and B 
respectively, and dropping subscripts and accents, 

x = — A cos 6 — B sin 6. 

If we consider this as a polar curve whose radius vector is laid 
off along the crank throw, we see that it represents a circle whose 
circumference passes through the pole, and whose diameter through 




the pole is inclined at an angle (270 — 8) with the reference line. 
For let O (Fig. 133) be the pole, and OQ the reference line. Draw 
OC = r and inclined at an angle (270 - 8) with OQ. On OC 
as a diameter describe a circle cutting the reference line at D, and 
cutting the perpendicular to it through O, at S. Then 



and 



(9^=-rsin8 = — A, 
OS=-rcosS = -B. 



VALVE GEARING 



185 



Draw any radius vector OQ inclined at an angle 6 with the refer- 
ence line. Draw CP' and DR' perpendicular to OQ. Then 
OP' = OR ' + R 'P' = — A cos — B sin = x. Hence the posi- 
tion of the valve with reference to its central position for any given 
angular position of the crank measured from its head-end dead 
point is given by the intercept OP'. The valve will be in its central 
position, or x will be zero when the crank is at OQ'. The valve 




Fig. 134 



will be at its maximum distance- to the left, i.e. will have its maxi- 
mum negative displacement, when the crank is at OC, and will 
have its greatest positive displacement when the crank is at OC. 
At the dead point O Q , x is equal to — A. This polar diagram 
is the same as was considered in connection with piston position, 
with the exception of the different phase angle. 

The above position of the polar locus or valve circle is correct 
when we take into account the algebraic sign of the valve displace- 
ment. But it is more convenient to draw another circle on CO 




Fig. 135 



produced to C, and consider the valve displacement as equal to 
OP, since, as we generally work with but one end of the cylinder 
at a time, the algebraic sign of the displacement is of but little 
consequence. By using the upper circle we have the advantage 
of measuring our radius vector in the direction of the crank throw. 
Let the valve shown by the dotted lines (Fig. 134) be in its 
central position, and let that shown by the full lines be the same 



1 86 



MECHANICS OF THE STEAM-ENGINE 



valve after having moved a distance x to the left, and opened the 
head-end port. In this second position call the amount that the 
steam port is open a x . Then, as x is negative, 




Fig. 136 



Now suppose that the valve moves a distance x to the right of 
its central position, so as to exhaust the port S. Then from Fig. 
135 we see that 

+ x = a 2 + i, 

a 2 = ( + x) — i, 

where a 2 is the amount that the port is open to the exhaust, and x 
is positive in this case. 

In Fig. 136'we draw a valve circle as in Fig. 133. We also draw 
a circle about O as a centre, and with radius equal to e, the outside 



VALVE GEARING 



87 



lap. Then 0P X = — x, 0R X = e, and R 1 P 1 = (— x) — e = a lt the 
amount by which the steam port is open for the position OQ of 
the crank. At the point X x where the valve circle intersects the 
e circle, — x = e, and a x = o. Hence at OQ x we have admission. 
When the crank has moved up to OQ , the dead point, the port is 
open the amount of the lead v. At Y lt — x = e again, and we 
have cut-off with the crank at OQ 2 . At OQ r the port is wide 
open by the amount FC\ When the crank has arrived at OQ' 
(the tangent to the circle at O), the valve will be in its central 




Fig. 137 



position, since there x = o. If it passes beyond OQ' we will 
have positive values of x, and positive intercepts on the circle OC. 
Let OQ (Fig. 137) be such a position of the crank. The value 
of -h x is OP 2 on the circle OC 1 . Now finally if we draw a circle 
about O and with a radius equal to i, the inside lap, then OR 2 = i, 
and R^P^ = (+ x ) — i = a 2 , the amount by which the exhaust port 
is open for any angular position of the crank. At OQ 3 , -\-x= i, 
and a 2 = o, so we have release. At O Q^ a 2 is again equal to 
zero, and compression begins, so that OQ 4 is called the Compres- 
sion Point. 



i88 



MECHANICS OF THE STEAM-ENGINE 



The regular polar or Zeuner valve diagram is shown in Fig. 138. 
Here an upper or negative circle is introduced as in Fig. 133, upon 
which — x can be laid off in the direction of the crank throw. It 
will be noticed that while the radius vector OQ is sweeping over 
the upper circle OC, the steam features of the head end will be 
taking place, and while sweeping over the lower circle OC the 
exhaust features of the head end occur. But it is evident also 




Fig. 138 



that the exhaust features of the crank end come within the circle 
OC, while the steam features of the same end come within the 
circle OC. These latter are, however, omitted in Fig. 138. The 
points <2i, (?2> (?3> an d <2 4 are the critical points of the cycle taking 
place in the head end of the cylinder. Variations of any of the 
quantities produce effects which are easily seen on the diagram. 
Increase in 8 admits steam earlier and cuts it off earlier. Increase 



VALVE GEARING 1 89 

in e admits steam later and cuts it off earlier. Increase in r admits 
steam earlier and cuts it off later, etc., and similarly for the exhaust 
side. 

Evidently neither a x nor a 2 can be greater than a, hence if r is 
greater than a + <?, the valve will overtravel the port. If in this 
case a circle be drawn about O as a centre, and of radius equal to 
a + e, its points of intersection U x and V 1} with the valve circle, 
will give the crank positions at which the port is just wide open. 
The distance o x on the diagram is called the Overtravel. Similarly 
o 2 is the overtravel on the exhaust side. The shaded portion of 
the diagram shows the variation in port opening. 

It is always well to make the valve as short as possible so that 
the pressure on its back and hence the work of moving it under 
this pressure may be a minimum. We must remember, however, 




Fig. 139 



that the ports are determined as to width by the general dimen- 
sions of the engine, and that the laps are also fixed by the design. 
The bridge b then should be as small as possible, but not so small 
that the valve will travel across it, and admit live steam into the 
exhaust port, as would be the case in Fig. 139. At the extreme 
travel of the valve x — r, hence in all cases we must have 

e + a + b > r, 

or b~> r — a — e. 

Neither should the exhaust port be so far covered by the inside 
edge of the valve as to make the effective opening of that port less 
than a. In every case then we must make 

#o + b — (2 + r) > a, 
or a > a + 1 ' + ** — b, 

as is evident from Fig. 140. 



190 



MECHANICS OF THE STEAM-ENGINE 



Problems in Valve Gearing. — In the design of any valve gear 
we have certain constants or dimensions given, and are required to 
find the values of all others. These constants may be any of the 
following : 

c s 
Constants of the gear 



e, i. 



Crank positions 



f <2n & Q 3 , or <2 4 -Oi or a 2 being zero.) 
Q in general. {a x or a 2 being known.) 



The last of these is the most important when Q = Q , and a x = v, 
the lead. 

It will be noticed that all four of the crank positions cannot be 
taken arbitrarily, as three will determine the position of a circle, 
and the fourth must follow from these. Also that one constant at 




Fig. 140 

least of the gear must be given, viz., e, i, r, or 8. If three crank 
positions are given, 8 is superfluous, and hence some other con- 
stant must be chosen. Also one at least of both steam and 
exhaust features must be given. The following are a few examples. 

1. Given e, i, y, and cut-off ; to find 8, r, Q a , and Q±. (y is 
the angle at which the crank stands when admission occurs.) Lay 
off y and the position of the crank at cut-off. (See Fig.. 138.) 
Draw the e- and /-circles. Draw the valve circle passing through 
the origin and the two points of intersection of the ^-circle with 
the given crank positions. The diameter of this through the 
origin will give r and 8. Produce this beyond O, and upon the 
prolongation draw another and similarly situated circle. The 
intersection of the /-circle with this last will give the release and 
compression positions. 

2. Given e, i, v, and cut-off position of the crank; to find the 



VALVE GEARING 191 

same quantities as before. Draw the e- and /-circles. Draw 
the crank positions at cut-off, and lay off the lead. Pass the 
valve circle through the origin, the lead point, and the point of 
intersection of the ^-circle with the cut-off position of the crank, 
and proceed as in problem No. 1. 

3. Given e y v, and r. Draw the ^-circle and lay off the lead. 
About the origin as a centre draw a circle of radius equal to r. 
Erect a perpendicular to the axis of X through the lead point, and 
where this intersects the circle last drawn will be the extremity of 
the diameter of the valve circle. 

4. Given cut-off position of the crank, a, and o x . (o x being the 
overtravel on the steam side.) We are also given the compression 
position or the release position. Call (# + o x ) — a'. We have 

e + a' = r, (1) 

x = r sin (8 + 6). 
If we put 6 = in this last equation, 

x Q = r sin 8 = e + v. . . . (2) 
If we put 6 = 2 where 2 is the angle of the crank at cut-off, 

x 1 = rsin(0 2 + $) = e, . . (3) 

we have three unknown quantities, viz., e, r, and 8, and three inde- 
pendent equations. Substituting for r in (1) its value in (2), we 

get 

e -f- v— {e + a') sin 8, 

a' sin 8 — v , N 

e= ir^T- • • • (4) 

Substituting in (3) the value of e from (4) and of r from (1), we 

g et 

'sin 8 — v fa' sin 8 — v . A . , n . «. 

+ a'\ sin (0 2 + 8), 



1 — sin 8 \ 1 — sin 8 
a' sin 8 — v = (a' — v) sin(0 2 + 8). 
Reduction gives 

sin 8 \a' + (v — a') cos 2 \ + cos 8 | (v — a') sin 2 ] = p. 
This is the equation of a circle in polar coordinates whose circum- 



192 



MECHANICS OF THE STEAM-ENGINE 



ference passes through the origin, and whose intercepts on the 
axes of X and Fare (v — a') sin 2 and a' -f- (v — a') cos 6 2 , respec- 




Fig. 141 



tively. Hence we lay off these known quantities on the axes (Fig. 

141), and pass a circle through the origin, and through their ex- 
tremities. Then we 
draw a second circle 
about the origin and 
with a radius equal 
to v. The line join- 
ing the point of in- 
tersection A of the 
two circles with the 
origin will give by 
its angle with the 
axis of X the value 
of 8. Since a' is 
always greater than 
v, and since sin 6 2 is 
positive in any prac- 
ical construction, the 
FlG . I42 term (v — a') sin 6 




VALVE GEARING 1 93 

will always be negative. Also it is readily seen that a + (v — a') 
cos 2 will always be positive. Hence the former is laid off to 
the left of the origin, and the latter upwards. The other inter- 
section B of the circles gives a perfectly correct though imprac- 
ticable value of 8. 

In solving the above problem a partly analytical and partly 
graphical method seems to be the best. Take the following : 
a 1 = J", v = -^q", cut-oif at \ stroke. To find 8, e, and r. If the 
cut-off comes at \ stroke, 2 = 6o° (approximately) . 

a'+ty-a 1 ) cos6> 2 = i+( T 3 g--i) cos 60°=^", 

(v - a') sin 2 = ( T \ - £> sin 6o° = l#». 

Now if we lay off (v — a') sin 2 vertically downwards from the 
origin, and a' +(v — a') cos 2 horizontally to the right, we will 
measure our 8 to the right of the axis of Fas it is usually placed. 
Hence the construction of Fig. 142 gives 8 immediately as 69 . 

Now e = a ' sin8 - v = - 2 M-' l8 7-=.6 4 ", 

1 — sin 8 1 — .933 

r — e + a! = .64 + .25 = .89. 

Some persons prefer a purely graphical construction for such a 
problem, even if more complicated in reality. Therefore the fol- 
lowing graphical method is introduced. About the origin as a 
centre draw a circle of radius OA = a' (Fig. 143), and one of ra- 
dius OV= v. Draw the given crank position at cut-off OQ 2 . Pro- 
duce Q 2 to M, so that UM=RO. Then OM=v-a\ and 
since angle A OR = 6> 2 , WO = {v -a') sin <9 2 , WM= (v-a') cos 6 2 , 
and by completing the parallelogram OMAB, we see that 
WB = WM+ OA =a' + (v—a') cos <9 2 . Hence the circle drawn 
on OB as a diameter gives by its intersection with the z'-circle the 
value of 8 correctly placed. 

In order to explain the theory of the construction for e and r, 
o 



194 



MECHANICS OF THE STEAM-ENGINE 



we must combine equations (i) and (2), remembering that 8 is 
now a known quantity. We have 

e + a' = r, 

e + v = r sin 8. 




Fig. 143 
Eliminating e from the above, there results 



r = 



a' — v 



If we divide both sides of (5) by a', we get 



a' — v 



a' a' — a' sin 8 

Hence from the point D, where OD cuts the ^-circle, drop the 
perpendicular DE. On ED lay off EF=VA=a'-v. Draw 



VALVE GEARING 



195 



AF, and produce to cut the axis of Fat G. Then will OG — r, 
for by geometry 

GO = FE 

A0~ AE 



or 



We now project G on the arc of a circle to C, and draw our valve 
circle on OC. The outside lap e will be GH— r— a'. 




Fig. 144 

Variable Cut-off with Plain Slide Valve. — If the point of cut- 
off is to be changed, it must be done by altering some of the con- 
stants of the driving gear, and not of the valve itself, since the laps 
and ports are absolutely fixed by the construction of the engine. 



196 



MECHANICS OF THE STEAM-ENGINE 



We must also be careful that by varying the cut-off we do not vary 
the admission position nor lead to any great extent. Let us first 
see how r and $ must be changed in order to give variable cut-off 
with constant lead. In this case, however the valve circle may 
change, all the circles must pass through the same lead point V, 
Fig. 144. Therefore the locus of the centres of all valve circles 
which give a constant lead will be a straight line at right angles to 
the middle point of the chord O V, and the locus of the extremi- 
ties of their diameters will be the perpendicular erected at V. 




Fig. 145 



In the figure are shown the positions of three valve circles, the 
extremities of the diameters being at A, B, and C. These all give 
the same lead EV, but the admissions vary slightly, being at Q x \ 
Qi', and Qi". The cut-offs vary widely through the range Q 3 ', 
Q 2 ", and Q 2 '". The exhaust features vary also. When the inside 
lap is considerable as shown, the release points do not vary much, 
being at Q 3 ', Q s ", and <2 3 '", but the compression points do, as is 
seen at Q 4 ', Q 4 ", and Q"'. If z = o, the release and compres- 
sion points vary an equal amount. It will be noticed that the 
change is produced by increasing 8 and diminishing r. It is 



VALVE GEARING 



197 



accomplished in practice by cutting a slot in the eccentric at right 
angles to the crank, so that it can be shifted across the shaft as in 
Fig. 145. This shifting is done automatically while the engine is 
running by means of a shaft governor. 

If we wish to keep the admission point constant, the circles must 
be varied as in Fig. 146. Here the locus of the extremity of the 
valve circle diameter is the perpendicular to the admission posi- 
tion OQ x at its intersection with the <?-circle. The three positions 




Q,Q,Q, 



A, B, and C give a constant admission at Q lf but a variable lead 
EV, EW, and EX. The cut-offs, releases, and compression 
points are similar to Fig. 144. This is the method of variation in 
most single valve, automatic, high-speed engines. The shifting of 
the eccentric at the dead point is as shown in Fig. 147. Since 
it is difficult to get the motion of pure translation given by the 
straight line ABC, it is usually approximated by means of the arc 
of a circle. 



198 MECHANICS OF THE STEAM-ENGINE 

Reversing Gears. — Valve gears which are so arranged as to run 
the engine in either direction are known as Reversing Gears. 
These must be applied to all marine, locomotive, and hoisting 
engines. Those reversing motions which are actuated by means 
of the so-called link are amongst the most interesting kinematic 
movements. By simply shifting the position of this very ingenious 
piece of mechanism, the engine may be run in either direction and 
with any degree of expansion. 

The Stephenson link is the one most generally used upon loco- 
motive and marine engines. Upon an axle O (Fig. 148) are keyed 




Fig. 147 

two eccentrics C and C, from which the rods extend and are 
joined to the two ends of the expansion link BB'. The block M 
on the end of the valve stem fits the slot of the link. The valve 
stem T is held by fixed guides, and the link can be raised and 
lowered at will. Let O (Fig. 149) be the centre of the engine 
shaft, and OQ Q the crank at its head-end dead point. In order 
that the engine may throw over, the eccentric must be set with an 
angular advance 8 as shown. If, however, the engine is to throw 
under as in Fig. 150, the angular advance must be 8 to the left of 
the perpendicular below the origin. Now if both eccentrics are 






VALVE GEARING 



199 



put on as in Fig. 151, and the ends of the arms are connected by 
the link, then when this is depressed the engine will throw over, 

























B y\ 


*^\(\ 


II c 


T 7 ^-^^ 


--^zzz: 


~j-^^Ll 


KSy 




-M 


\ \\^ c 








5Tp 


T 










: ^^**jJI 







Fig. 148 

and when raised it will throw under. In these cases one eccentric 
will control the motion of the valve. When the link is in any 
other than one of its two extreme positions, the motion of the valve 
will partake of the motion of both eccentrics. 




Fig. 149 



By considering the length of the eccentric rods very great, and 
by making the radius of curvature of the link very great also {i.e. 



200 



MECHANICS OF THE STEAM-ENGINE 



by making the link straight) , we will have the arrangement shown 
in Fig. 152. Here the points of attachment BB' (Fig. 148) are 
supposed to lie in the pitch line of the link, and that of the block 
M remains rigidly fixed in the link for any given position. When 




Fig. 150 

the crank is on the dead point OQ , the eccentrics are at 0C X and 
OCi, and the link at B x By. When the crank is at the other dead 
point, the eccentrics are at 0C 2 and 0C 2 , and the link at B 2 B 2 . 
XYis the mean of these two link positions, hence we can measure 
movements of points of the link from it as a central position, and 
as the valve is rigidly attached to the link block, these movements 
will correspond exactly to valve movements. Let the crank turn 




Fig. 151 

through an angle 0, and consider the motion of the extremities of 

the link which have now moved to B and B'. Let BS equal x lt 

and B'T equal x 2 . Since x x is entirely due to eccentric C, we 

have 

x x =r sin (0 + S). 



VALVE GEARING 201 

And for x 2 , the motion of the other end of the link, we have 

x 2 = r sin (0 + (180 — 8)). 
Expansion of these two equations gives 

x x = A cos 6 -+- B sin 0, 

x 2 = A cos — B sin 0. 

Now consider the motion of any other point of the link such as 
M, and call the distance MN = x. Let the whole length BB' 




Fig. 152 

of the link be represented by a, and the amount B'M by Ka. 
Then by similarity of triangles, 

x 1 — x 2 : x — x 2 : : a : Ka : : 1 : K. 

x = Kx x -+- (1 — K)x 2 . 

Substituting the values of x 1 and x 2 , 

x = Ka cos O + KB sin + (i—K)A cos 6- (1 -K)B sin 0, 

x — A cos 6-\-(2K — i)B sin 0. 

This is the equation in polar coordinates of a circle passing 
through the origin whose intercept on the axis of X is A, and 
on the axis of Y is (2 K — 1) B. It will be observed that how- 



202 



MECHANICS OF THE STEAM-ENGINE 



ever the link be moved, that is, however K be varied, the intercept 
A will not be changed, or the lead will be constant. The system 
of valve circles is shown in Fig. 153. OC is the valve circle 
corresponding to the upper eccentric, and OC that of the lower 
one. The angle/0C=8, and the angle JO C ' = 1 8o° - 8, 
OV=A, the same for all the circles OJ=B, and OJ'= —B. 
If any radius vector be drawn as O Q, OP Y = x lf OP 2 = x %- If 



J 


>- — -\c 


A! 






<-\r-zA-y>\ \ 


J' 


^ 'Z^Q 



Fig. 153 



we lay off OH= (2K— 1) B, and pass a circle through O, V, and 
H, we have circle OZ, representing the motion of the valve for any 
intermediate position of the link, and OP=x. 

For rods of finite length the action is far more complex. Let 
CC (Fig. 154) be the position of the eccentrics when the crank 
is on the head-end dead point. Let the pitch curve of the link 
slot be the arc of a circle whose radius is in general R. Then 
with the crank stationary, the point B will, as the link is raised 



VALVE GEARING 



203 



and lowered, follow the circle PA about C as a centre. Let 
BqSPq be the link at mid gear, i.e. when B Q and B ' are equi- 
distant from the line of connection. It is readily seen now that 
the lead cannot be constant, as the extent of its variation, or the 
difference between the maximum and minimum widths of the lead 
will be PS. When the crank is on the crank-end dead point, the 
arms of the link will be crossed, and the locus of B, as the link is 
raised and lowered, will be the circle A'P' about C 2 as a centre, 
and the variation in lead will be S'P'. The best that can be 
done, then, is to so choose the radius of the slot that the variation 
of the lead on the two ends will be the same. Let us find, then, 
an expression for the distance PS. 

a\ 4 











L 




M 


1 V 


/ ~ \ 

/ D \ 


--B L '-' 


R__ 





pi 


T J 

's rl 




I 0) 




-J ^J^^ 






r ( 


u 


;. 


C"^- 


—<v 










T1 

1 
1 





Fig. 154 



When the link is in mid gear the centre of curvature of the slot 
will lie on the axis of X, as the whole apparatus is then symmet- 
rically arranged about that axis. Let it be at D, and call the 
distance OD = b. Then 

OS=fi + P. 

Call the length of the chord B B ' = a, a constant of the link, and 
denote the point at which this chord cuts the axis of X at right 
angles when the link is in mid gear by 7. 



{OT-b) 2 + - = R\ 
4 



b=OT-^P 2 -- 



204 MECHANICS OF THE STEAM-ENGINE 

But OT=MG==MC-CG=\L 2 -f~-rcosS\ 2 -rsm8, 
where Z is the length of the eccentric rod or link blade. 



=V" 



Hence b=\D- f -- rcosS ) -rsinS-\R 2 -- 
and finally, 



„^s Jm « 2 . 



\L 2 - f- - r cos sY - r sin 8 -\R 2 - - + R. 



OS. 

2 



Now the distance OP we know to be 



0P= Z V i - n 2 cos 2 3 - r sin 8, 



where n='y Hence the variation in lead is 



PS = OS- OP, 

PS = \L 2 -(--r cos Z)-\R 2 --+R- ZVi-« 2 cos 2 8. 



At the other dead point we obtain an expression for OS' by 
putting i8o° — 8 for 8; hence 

OS' =aJl 2 -(- + r cos sY + r sin 8 -V^ 2 - - + #> 
■ \ 2 7 % 4 

and OP' = ZVi-« 2 cos 2 8 + r sin 8. 

^'5' = OP' - OS', 



p>S' = -\L 2 -(- + rcos8)\\R 2 ---R + LVi-n 2 cos 2 8. 



* J " 4 

But these distances are to be equal ; hence 



xM 



4 



= ^Z 2 -^-rcos8Y+i^/z 2 -^ + rcos8Y-ZVi-« 2 cos 2 8. 



VALVE GEARING 



205 



The right side of the equation contains constants only, and there- 
fore can be calculated for any given case. Call it C. Then 



R = 



2 C 



From this we may obtain an exact result for the radius of the link 
slot. However, as r is usually small in comparison with L and a, 
we may expand the first two radicals by the binomial theorem, 
then add them term for term, and reject all terms containing 
the square or higher powers of r. The remaining terms will 

be merely the expansion of I D ) • Likewise we may neglect 

the term n 2 cos 2 8, n 2 being too small to be considered. The 
approximate form of the equation then is 



V*-J-*-V^?-A 



which is satisfied when R = L. 

In most cases it is amply close enough to make the radius 
of the link slot equal to the length of the rod or blade. When 




the link is not hung on its pitch line, the radius of the slot will be 
practically the same. 



206 



MECHANICS OF THE STEAM-ENGINE 



In order to investigate the steam distribution, graphical methods 
can be best used. Let the crank stand at any angular position 
(Fig. 155), and lay in the eccentrics C x , C\', in their proper 
positions. Having the length L of the blade given, draw the 
locus circles AP oi the point B, and A'P' of the point B'. Con- 
struct a template of thin sheet metal as shown in the figure, the 




Fig. 156 

corners at B and B' representing the same points as in Fig. 148, 
and let the edge SS' be drawn with the proper radius R. The 
distance BB' is, of course, a. The link is suspended from some 
point on its pitch line (usually the middle point U) by a radius 
bar from a point V. This point V can be shifted along an arc, 
thus varying the steam distribution. Suppose in the present 
instance the link is suspended from the point V x . Then with 



VALVE GEARING 



20? 



V x as a centre, and a radius equal to the length of the radius bar, 
strike an arc U x . Now shift the template upon the drawing, keep- 
ing B and B' over their respective loci, until the middle point U 
of SS' falls upon the circular arc UU X , and draw in the arc SS 1 . 
Then will the distance x, measured from the middle line XX to 
P x , be the distance that the valve is drawn past its central position. 
The line XX is obtained by taking the mean of the positions of 
P x at mid gear for the two dead points. Now take a new position 
of V, say at V 2 , strike the arc C7 2 , shift the template till U falls 
upon this, and measure x again. Having obtained a number 



Fixed Point 




FIG. 157 



of such measurements for different positions of the link, proceed 
to lay them off on a Zeuner diagram as intercepts Oi, O2, 0$, 
etc., along the crank throw OQ in its given position 6. Now take 
a new crank position and go through the same construction again, 
using the same points V 1} V. 2 , V S) etc., as before. By connecting 
corresponding points on the Zeuner diagram with a smooth curve, 
an exact polar diagram of valve travel can be obtained. If the 
length of the link blade is sufficiently great to admit of the motion 
of the valve being considered as simply harmonic, the above curves 
will be circles, and but two positions of the crank will be necessary 



208 



MECHANICS OF THE STEAM-ENGINE 



to determine all the circles. These two positions may conveniently 
be 6 = o°, and = 90 , as shown in Fig. 156. 

If the blades are crossed with the crank on the crank- end dead 
point as shown above, the link is known as a crossed link, and 
the lead will be least at mid gear. If the blades are crossed on 
the head-end dead point, the link is an open link, and the lead 
will be greatest at mid gear. 

In the Gooch link the link itself is not raised and lowered, but 
the valve rod is (Fig. 157). It has the property of giving an abso- 
lutely constant lead, but otherwise the steam distribution is not as 
good as that given by the Stephenson link. It requires more room 
also to get in the long arm AB. 



{fi) The Gridiron Valve 

In this valve gear the steam chest is divided into two compart- 
ments by means of a partition parallel to the valve face. In the 
lower compartment the distribution valve works. It is a plain 
slide valve with small laps. In the upper compartment is the 
expansion valve, which consists of a plate with a rectangular 
opening in the centre. This plate slides back and forth over a 

single port in its 

seat, and cuts off 
the steam with its 
inner edges. The 
motion of this valve 
will be a simple 
harmonic motion, 
and will be given 
by an equation of 
the same form as that of the plain slide valve cutting off with its 
outer edges. If we represent by x the distance which the ex- 
pansion valve has moved past its central position, for any given 
angular position of the crank, by r the throw, and by S the 
angular advance of the eccentric which drives it, our equation of 
motion will be of the form 

^ =r sin(^ + 8 ). 




Fig. 158 



VALVE GEARING 



209 



The eccentric will, however, usually have a negative angular advance. 
Now consider the expansion valve in its central position as shown 
by the dotted lines in Fig. 158. Call the total width of the port 
a , and the amount by which the edge of the valve is beyond the 




Fig. 159 

edge of the port e . Now let the valve move a distance x to the 
left, throttling up the port to width a x . Then we will have 

x -\-a 1 = a + e , 

a 1 = (a + e ) — x . 

The value of x can be found at once by means of a Zeuner 
diagram. In Fig. 159, OC is the expansion valve circle, inclined 
at an angle — 8 to the left of the perpendicular to OQ . OQ is 



2IO 



MECHANICS OF THE STEAM-ENGINE 



any position of the crank, and hence OP Q = x . With centre O 
and with radius a + e 0> describe a circle. Then OP = a + e Qt 
and P R = (a + e ) — x = a x . We can conveniently measure 
the port opening as affected by the left hand edge of the valve 
by means of another circle drawn on C O produced below the 
origin. The expansion valve will then admit steam into the lower 




Fig. 160 



steam chest by its left edge when the crank is at OQi and cut it 
off with its right edge at O Q 2 '. If we draw the distribution valve 
circle on the same diagram, we can follow the whole distribution 
of the steam. In Fig. 160, OC is the distribution valve circle at 
its angular advance 8, and OC the expansion valve circle at its 
angular advance 8 . OR — e, the outside lap of the main valve, 
and OR — a + e of the expansion valve. For any crank posi- 



VALVE GEARING 211 

tion OQ, /y? is the port opening of the expansion valve, and 
PR the port opening of the distribution valve. The expansion 
valve opens first at OQ x l , but steam does not flow into the cylinder 
till the distribution valve opens at OQ x . The expansion valve 
closes first at OQ 2 ', and hence determines the cut-off. The main 
valve closes at OQ 2 before the expansion valve opens again. If 



Fig. 161 

the distribution valve fails to close before the expansion valve 
opens again, there will be double admission. It is evident that if 
#o + e > r o> there will be no cut-off by the expansion valve. The 
shading shows the effective port opening. 

This form of expansion valve is frequently made with a number 
of slots or openings in the valve, and a corresponding number of 
ports in the seat (Fig. 161) ; hence the name "Gridiron Valve." 

(e) The Meyer Valve 

This gear produces an early cut-off by means of an expansion 
valve running on the back of the distribution valve. The distri- 
bution valve consists of a flat plate (Fig. 162), in which are cut 
two rectangular ports D, Z>, as well as the exhaust hollow. It is 
evident that its action is exactly the same as an ordinary slide 
valve. The expansion valve runs on top, covering the ports D, D, 
and thus effecting the cut-off. If the distribution valve is driven 
by an eccentric of throw r and angular advance 8, and the expan- 
sion valve by one of throw r and angular advance 8 , we will have 
as before 

*=rsin(0-}-S), 

x = r sin (0 + 8 ) • 

Now x and x are the motions of the valve referred to the fixed 
valve seat ; but the cut-off is determined by the relative motion 



212 



MECHANICS OF THE STEAM-ENGINE 



of the two valves, or by (x — x ). Let us first get the equation 
of relative motion. The upper portion of Fig. 162 represents the 
two valves in their central positions (a state of affairs which could 
never occur when the engine is running, however) . The lower half 
shows the valves after having moved to the right ; the distribution 
through a distance x and the expansion through a distance x . Uy 




Fig. 162 

is the distance from the edge of the expansion valve to the outer 
edge of the port D, when the upper valve is in its central position 
relatively to the lower one, y will be a constant of the valve gear. 
Call the port opening of D a x as before. The equation of motion 
is evidently 

a x + x =y + x , 

a 1 =y — (x—x Q ). 



If we draw two valve circles representing x and x , as shown in 
Fig. 163, OP=x, and OP = x . Hence PP = x — x . But 
this last distance is not measured from a fixed point, viz. the 



VALVE GEARING 



213 



origin, and hence we cannot subtract it graphically from a con- 
stant quantity y by drawing a circle of radius y about the origin. 
We must first, therefore, find the curve which will give for every 
position of the crank the value of x — x measured from the 
origin. 

x — x = r sin (6-}- 8) — r sin (0 + 8 ) = z, 

z = r sin 8 cos 6 + r sin 6 cos 8 — r sin 8 cos — r Q sin cos 8 , 

2 = (r sin 8 — r sin 8 ) cos + (r cos 8 — r cos 8 ) sin 6 

= A'cosO + £'smO, 

where A' and £' are constants of the valve gear. This is the 
equation of a circle whose intercepts on the axes of X and Y are 




Fig. 163 



A' and B' respectively. In Fig. 164 let OC= r and COY=S; 
let OC = r and C OY= 8 . Connect C C, and complete the 



214 



MECHANICS OF THE STEAM-ENGINE 



parallelogram OC CC Z , thus determining the side OC z . From 
C , C, and C z drop perpendiculars on the axes of X and K 
Then 

SO = XC = OH- <9£ = rsinS-r sinSo = ^', 



OT=KC= CH- C G = rcos8-r cos 



B' 



Hence the circle on OC z as a diameter will give by its intercepts 
on the axes of X and Y the values of A and B\ and by its 
intercept on any radius vector {x — x ) for that particular crank 
position. Fig. 165 shows the complete Zeuner diagram for the 
Meyer valve. OC is the throw of the distribution valve eccentric 
with its angular advance 8. OC is that of the expansion valve. 







Fig. 164 



From these, by completing the parallelogram, we get OC z and S z 
of the auxiliary circle. On OC and OC z as diameters, describe 
circles. There is no use in drawing a circle on OC , as it is not 
used. Then if OQ is any position of the crank, OP=x and 
OP z = x — x = z. We now draw a jy-circle about the origin, 
hence OR z —y, and P z P z =y — z = a 1 . The expansion valve 



VALVE GEARING 



215 



opens first at Qi, but steam is first admitted into the cylinder 
at OQ x , where the distribution valve opens. Cut-off is determined 
by the expansion valve at OQJ, and the distribution valve closes 
at OQ 2 , before the expansion valve opens again at OQi. The 
second opening of the expansion valve on Qi'O produced is on the 
other end of the valve, and hence does not interfere with OQ 2 in 




Fig. 165 



such a way as to cause a double admission on the head end. Cut- 
off can be varied by varying the value of y. This can be done 
by making the cut-off plate in two parts, which can be separated 
by means of a right and left hand screw on the valve stem. Another 
way is by varying 8 with an automatic shaft governor, as in Fig. 
166. 



2l6 



MECHANICS OF THE STEAM-ENGINE 



(a 7 ) The Thompson Valve 

In this valve gear, which is used on the Buckeye engine, the 
arrangement of valves is the same as in the Meyer gear. The 
driving mechanism, however, is different. The distribution valve is 
driven directly by an eccentric r (Fig. 167). The point B, where 




Fig. 166 



the eccentric rod is hinged to the valve stem, is carried on a 
vibrating pillar BC. The expansion valve is driven from the 
extremity A of a lever AE, whose middle point D is pivoted to 
the middle point of BC, and whose other end is driven by an 
eccentric r z . We have, as in the case of the Meyer valve, 

a^y-ipc-x^), 



VALVE GEARING 



217 



where a x is the port opening, y a constant, and x and x distances 
moved through by the distribution and expansion valves, both 
measured from their central positions referred to the fixed valve 




Fig. 167 

seat. The value of x will be determined both by r and by r z . 
Since AB — BD and DC = DE, that component of A J s motion 
due to r will be simply x, where 

x = r sin (6 -f 8) . 

That component of A's motion due to r z will be — x z , and 

-* 2 = -^sin(0 + S,). 

Now x is the algebraic sum of these two as they take place in the 
same straight line. Hence, 

a x =y — (x — x + x z ) =y — x z . 

The motion of the expansion valve relatively to the distribution 
valve when driven through the above kinematic chain will be the 
same as if the expansion valve were driven directly from the eccen- 
tric r z upon a fixed seat. The form of the equation of motion is 
identical with that of the gridiron valve, hence the travel of the 
expansion valve on the top of the distribution valve will be of con- 
stant magnitude for all cut-offs when the expansion eccentric is 
rotated about the shaft by means of a shaft governor. 



218 



MECHANICS OF THE STEAM-ENGINE 



B. Relation between Position of the Crank and Veloci- 
ties and Accelerations of the Valve 

Since the motion of the valve is taken as simply harmonic, we 
can express the velocity of the valve for any angular position of 
the crank by differentiating xj with respect to time. 

x B ' = — r sin (8 + 0), 

dxj ,« , n ,d0 

-^=-,003(8 + *)-. 



And if — = (o = constant, 
dt 




Fig. i68 

If this be laid off as a polar curve along the crank throw as a 
radius vector, we see that it represents a circle (Fig. 168), whose 
diameter is rw, whose circumference passes through the pole, and 
whose diameter through the pole is inclined at an angle i8o°— 8 
with the axis of X. 



RELATION BETWEEN PISTON AND VALVE POSITIONS 219 

The acceleration of the valve would be obtained by differentiat- 
ing with respect to time the expression for the velocity, or 

at at 

= ro> 2 sin(S + 0). 
Y 




Fig. 169 

This is the equation of a circle (Fig. 169) whose diameter is roi 2 , 
whose circumference passes through the pole, and whose diameter 
through the pole makes an angle of (90 — 8) with the axis of X. 



4. RELATION BETWEEN PISTON AND VALVE POSITIONS 

In the two preceding sections we have obtained exact relations 
between the crank and piston position and the crank and valve 
position. It is the purpose of the present section to investigate 
the relation between the position of the piston and valve. Analytic 
expressions for this relation are, in the exact case, and even in the 



220 



MECHANICS OF THE STEAM-ENGINE 



approximate ones, too cumbersome to be employed. But by the 
graphic line of analysis already discussed the solution is direct and 
complete. 

Let Fig. 170 represent a polar valve diagram. About O 
describe any circle, as that through Q, to represent the crank orbit. 
If OQ is any position of the crank, PR* is the port opening. But 
the piston position corresponding to Q on the diameter of the 
crank circle as a stroke is at E, projected on a circle about Pas a 
centre. Hence piston position E corresponds to port opening 
PR 1 of the valve, and piston positions E 1} E 2 , E s , and E 4 corre- 
spond exactly to admission, cut-off, release, and compression. 



3f 








/ 


QjZjf 




e 
i 7 


2h — ____ 


~ — — — . p 


vf\ 


\p 









Fig. 170 



Conversely, if any three of these piston positions be given, a valve 
can be designed to fulfil the conditions. 

If the laps are the same on the two ends, then evidently the pis- 
ton positions corresponding to admission, cut-off, etc., cannot be 
the same on the two ends. Consider, for example, the cut-off 
positions for equal laps in Fig. 171. The crank positions OQ 2 ' 
and OQ 2 " will be diametrically opposite, but the piston when at 
E 2 ' will have completed a larger fraction of its stroke than when 
at E 2 ". This defect can be remedied by making the outside laps 
different on the two ends, giving the larger to the head end, but 
by doing so the admission and lead on that end are changed. For 
the same reason the head end will have a greater compression, 



RELATION BETWEEN PISTON AND VALVE POSITIONS 221 

as the exhaust will close when the piston is farther from the end of 
the stroke. If a valve is set with equal leads, the cut-off will be 
later and the compression will begin earlier on the head than on 




Fig. 171 



the crank end. With a variable cut-off, such as is produced by a 
shaft governor (Figs. 144 and 146), or by a Stephenson link 
(Fig. 156), if the cut-offs be equalized in one position by change 
of laps, they will not be exactly equal for any other position. 



CHAPTER II 

DYNAMICS 

The forces acting upon the moving parts of the steam-engine 
are due to two causes: (i) to steam pressure, and (2) to the 
inertia effects of the moving masses. Of these two the first is in 
general the largest and most important, and it is the force which 
does work outside of the machine itself. The forces due to the 
second cause are resident entirely within the machine, and the net 
work due to them within any complete revolution made under 
steady conditions is always zero. We will take up the discussion 
of these forces in the order given. 

1. FORCES DUE TO STEAM PRESSURE IN THE CYLINDER 

These forces are resident entirely within the cylinder of the 
engine, and they will act upon the piston in the direction of the 
main line of connection of the machine, that is, in the direction of 
the motion of the piston. Their magnitude then is the only im- 
portant consideration, and this can best be studied by reference 
to a curve in rectangular coordinates, which shows the relation 
between piston position and total steam force. This curve can be 
deduced from the familiar indicator card, and for the accurate 
study of the steam effects it is best to use an actual card, either 
taken upon the engine under discussion or upon another of the 
same type. If no such card is at hand, an approximation to its 
form can be drawn after the following constants and dimension? 
are known : 

1. The boiler pressure. 4. The release point. 

2. The back pressure. 5. The point of compression. 

3. The point of cut-off. 6. The clearance. 



FORCES DUE TO STEAM PRESSURE 



223 



Of these, 1 and 2 are simply working conditions ; 2 is immedi- 
ately known in the case of the simple engine, but must be obtained 
by calculation in the case of the compound ; 3 and 5 are known 
from the valve design, and 4 may be taken as occurring at the end 
of the stroke without serious error ; 6 may be assumed from what 
is known to be an average value in the particular type of engine 
treated ; 3, 5, and 6 can be expressed as fractions of the stroke 
or of the volume swept through by the piston. 



B' 


D' 




J' 
C 


^-^7' 




F' 

G' 








1/ 


P L 


H E 


G 

F 






__J 


J 








Y 


^ 


N 
\ 





D 



B 



Fig. 172 



The Simple Engine. — The cycle of changes of pressures and 
volumes taking place upon one side of the piston during one revo- 
lution of the engine is as follows : steam is admitted into the head 
end of the cylinder at boiler pressure (or a little less), at B (Fig. 
172), and as the force exerted by the steam upon the piston is 
toward the origin O, or to the left, this pressure is negative and is 
laid off below the axis of X. Call this total pressure, i.e. the spe- 
cific pressure multiplied by the area of the piston, JP lt and we must 
measure it above the vacuum or line of zero pressure. Hence 



224 MECHANICS OF THE STEAM-ENGINE 

BH=P X . This pressure is maintained pretty constantly up to the 
point of cut-off at D, so that DL = P 1 also. Let BE = AB = s c) 
the clearance expressed in terms of the stroke BC=s of the 
piston. At D expansion begins, the pressure dropping off on a 
curve DYE, which is nearly enough a rectangular hyperbola with 
asymptotes EA and EC to be taken as such by the designer. 
The terminal pressure CE=P 2 is determined by the equation of 
the rectangular hyperbola, viz. 

■^2— ; ■> 

s +s c 

or by graphical construction. The release then occurs, the press- 
ure dropping still farther to CG = P 3 , the back pressure, which 
in a non-condensing engine is about 17 lbs., while in a con- 
densing engine it is from 6 to 8 lbs., multiplied by the piston 
area. This back pressure is maintained at a fairly constant value 
during the return stroke until at / the exhaust closes, and compres- 
sion begins, the pressure rising on the hyperbola IJ to P c , where 
P c is given by 

x c 

Meanwhile a similar cycle of pressures has been taking place on 
the crank end, and as all forces acting on the piston due to steam 
in this end are positive, the curves will all lie above the axis of X. 

The work done by the steam in the head end in passing from 
B to C will evidently be proportional to the area BDFCH, and 
the work done upon the steam in the head end in passing from C 
back to B will be proportional to the area GIJHC. Hence the 
total or net work done in the head end during one complete revo- 
lution is proportional to the area BDFGIJ. Similarly on the 
crank end of the cylinder the net work done in one complete revo- 
lution will be proportional to the area B'D'E' G'f'J', and the whole 
output of work will be proportional to the sum of these areas. 

Now as the piston passes from B tc C, the negative forces will 
be given by the ordinates of the steam curve of the head end 
or by BZ>E, and during the same stroke the positive forces are 



FORCES DUE TO STEAM PRESSURE 



225 



given by the ordinates of the back pressure curve of the crank 
end. Hence the net force acting on the piston-rod will be given 
J 





^~-^___T Position Pressure 


G' 


z 


. 


H 


c 


\ 




F 








Nv\F 


X 




D 


B 



Fig. 173 

by the algebraic sum of these curves or by the ordinates of XYZ 
( Fi g- T 73)- Th e area XYZCH will be proportional to the net 
work done by the steam in passing from H to C, or during one 
forward stroke of the engine. Similarly the area under the curve 



B' 


D' 






X 




\ 


















F J 


c 








H 




Negative Pressure 


1 
Z 


< 


? 




/^\ 


J 



Fig. 174 

X'V'Z'UC (Fig. 174) is proportional to the work done by the 
steam during one backward stroke of the engine. These areas will 
Q 



226 MECHANICS OF THE STEAM-ENGINE 

not in general be equal to the areas of the original cards BDFGIJ 
etc., unless the cards themselves are exactly alike, but it is evident 
on comparison of areas that the sum of the areas of the curves 
XYZCH and X'Y'Z'HC will be equal to the sum of the areas of 
the indicator cards. 

These last curves of Net Horizontal Steam Effort are particu- 
larly useful in the analyses which follow. 

The Multiple-cylinder Engine. — In a multiple-cylinder engine 
the steam is expanded through more than one cylinder, the first 
taking its steam from the boiler, the second taking it from the 
exhaust of the first, and so on. In every case all the cylinders 
work upon the same shaft and have the same stroke, though the 
arrangement of the cranks may be different in different cases. It 
is readily seen that the steam line of any cylinder beyond the first 
up to its point of cut-off will be identical as to specific pressure at 
any given instant of time with the back pressure line of the pre- 
ceding cylinder. Here, as before, it is best to base the analysis of 
the acting forces upon actual indicator cards. But if unobtain- 
able, theoretical ones must be drawn. It is not the province of 
the present work to go into the calculations necessary to obtain 
the pressures around the cycles of a multiple-cylinder engine, that 
being more a problem in thermodynamics. It is sufficient to say 
that such cycles can be drawn quite accurately to scale when the 
proper dimensions and constants are known. The curves of any 
one cylinder can then be combined to find the net horizontal force, 
as in the case of the simple engine. When the several pistons act 
upon a single rod and crank, in other words are tandem, the net 
horizontal pressure curves can be added directly. But when act- 
ing on different cranks each cylinder must be considered by itself. 

Power developed in the Cylinder. — The power developed in 

the cylinder can readily be calculated from the net pressure 

curves. Let A be the area under the curve XYZCH, obtained by 

a planimeter in square inches, and b the length HC in inches. 

A 
Then — will give a number which, on the scale of forces used, is 
b 

the total mean pressure. This constant force acting on the piston 



FORCES DUE TO STEAM PRESSURE 



227 



would do the same work in one stroke as the variable force before 
considered. Call it P m , and let PJ be a similar pressure for the 
backward stroke. Then 

P m x (number of forward strokes per minute) x L 
33000 

PJ x (number of backward strokes per minute) x L 



+ 



H.P., 



33000 

where L is the length of the stroke in feet, and H.P. the horse- 
power developed. 



H.P.= 



(P m + PJ)xZxN 



33000 

N being the number of revolutions per minute. 

In all the above the pressures are taken as total pressures act- 
ing on the piston-rod. If specific pressures are employed, the 
area of the piston will enter as a constant factor on the right side 
of the equation. 

Forces at the Wrist Pin. — The net horizontal force active on 
the piston-rod being now known, its components at the wrist pin 
can be obtained. In Fig. 175 P x is this horizontal force. Its 

T 
Q z 




Fig. 175 

component at right angles to OX will be P y , which is the thrust 
upon the guide. Its component along the connecting rod is P x . 
P y can be plotted along the path of P as a curve of normal press- 
ures. As the force is at right angles to the direction of motion, 



228 



MECHANICS OF THE STEAM-ENGINE 



the area of the curve represents no energy. The general form of 
the curve is shown in Fig. 175. On the return stroke from C to 
H a similar curve will be produced. 

Forces at the Crank Pin. — The force P z will be transmitted 
directly to Q as a force Q z . This can be resolved in the tan- 

















\ 




0' 




Q 


H' 



Fig. 176 

gent and normal to the crank orbit, giving components J 7 and 
N. The tangential component is the true turning force of the 
main shaft at O. If we develop the semicircumference I/'QC 
along a straight line, as in Fig. 1 76, and lay off along this as a 
base a curve whose ordinates are the tangential forces T, the 
area of the curve will be proportional to the total work done 
upon the crank pin during a semirevolution from H l to C, as 
the force ^acts in the direction of the motion of Q. In other 




Fig. 177 

words, the area of the curve H'TCB' (Fig. 176) must be equal 
to the area XYZCH (Fig. 173), when the two are drawn on the 
same scale of pressures. If now we divide the area of this 
curve in square inches by the length of its base in inches, we will 
get a number which, on the scale of forces employed, will be the 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 229 

mean turning effort. If this constant force be applied at right 

angles to the crank, it would do the same work in passing from 

H [ to C as the variable force previously considered. Call this 

A A 

mean turning effort T m . Then — = T m . But - = P m , and b = 2/. 
TT 7r/ b 

Hence, 

T m = -P m . 

7T 

The normal component at the crank pin may also be laid off as 
the ordinates of a curve along the developed semicircumference 
as a base, giving results something like Fig. 177. The area of 
this curve represents, no energy, as the direction of N is at right 
angles to the path of Q. 



2. THE INERTIA EFFECTS OF THE RECIPROCATING PARTS 

A. Analytical and Graphical Calculation of Forces Active 
at the Crank and Wrist Pins 

The method rests primarily upon two principles of Analytic 
Mechanics, viz.: (1) that the sum of all forces acting upon a 
body when resolved in a given direction is equal to the mass of 
the body multiplied into the acceleration of its centre of mass in 
the given direction and (2) that the moment of all forces about 
an axis through the centre of mass is equal to the moment of 
inertia of the body about that axis into its angular acceleration. 

In order to clearly illustrate the application of these principles 
to the question in hand, let us consider the case exhibited in Fig. 
1 78. Here O is the centre of a shaft of a horizontal steam-engine 
and OQ is the crank, turning with a constant angular velocity w. 
QP is the connecting rod with wrist pin at P. Let / represent the 
throw of the crank, L the length of the connecting rod measured 
from centre to centre of brasses, and r the distance of the centre 
of mass G of the rod from the centre of the wrist pin. Taking O 
as the origin of coordinates, and the line of connection of the 
engine as the axis of X, we will call the coordinates of G x and y. 
Let be the angle between any position of the crank and the axis 



230 



MECHANICS OF THE STEAM-ENGINE 



of X, and let a be the corresponding angle of the connecting rod. 
Q x and Q y are the component forces acting on the crank-pin end 




Fig. 178 



of the rod parallel to the axes of X and Y respectively, and P x and, 
P y are the components at the wrist-pin end. Then if M' is the 
mass of the rod, we will have from the first principle 



Q X + P X = M 



1 (Px 



Q, + P„ = M'^ + M'g. 



(x) 
(2) 



Projecting these forces at right angles to the rod, and taking the 
moments about G, we get from the second principle 

— .{ — Qy cos a + Q x sin a\ (L — r) + \P X sin a — P y cos a] • r 

d 2 a 



A 



dfi 



(3) 



In these equations P x is any external force acting on the wrist- 
pin end of the rod parallel to the axis of X, and is due to steam 
pressure on the piston, the acceleration of the cross-head and 
piston, and to friction. It may be either positive or negative, but 
can always be calculated for any value of 0. I is the moment of 
inertia of the rod about an axis through G parallel to the main 
shaft, and can be experimentally determined. We have, there- 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 23 1 

fore, three equations from which to determine three unknown 
quantities, — Q x , Q y , and P y . Solving (3) for P y , we get 

Iq — - + (L — r) Q x sin a — (L— r) Q y cos a — r P x sin a 

p= * — 

y r cos a 

Substituting this in (2), 

79 

I — - -f- (L — r) Q x sin a — (Z — ;-) Q y cos a — rP x sin a 
dr 

r cos a 

Substituting for Q x its value from (1), and solving for Q y , we 
finally get 

Q s= J^^ + L M ,g + LM- g 

L cos a at- L dt~ L 

+£—? M } **** tan u-P x tan «, . . (4) 
L dt" 

Q X =M'^-P„ (5) 

dt~ 

P^M^-M'g-Q, (6) 

In these three equations there are, in addition to the unknown 
quantities, certain dimensions and constants of the engine as well 
as three accelerations which must be determined. These latter 
we have already deduced in Chapter I. Since G lies in the axis 
of the rod, 

d*x /9 n 9/7 x (cos 2 4- ?r sin 4 0) ,„v 

— - = — /orcosfl — «(!)-(/— nr)- — -, . . (7) 

dt " (T-/r'sin 2 0)* 

-t4 = —rn or sin 0, . (8) 

dr x 



<d 2 c t _ na) 2 (i — n 2 ) sin 
dt 2 (i_^ sin 2 0)f 



(9) 



232 MECHANICS OF THE STEAM-ENGINE 

It is also convenient to write down, 



Z cos a = ZVi — n 2 sin- 0, 

n sin0 

tan a = — — • 

Vi-« 2 sin 2 

All of these are known in terms of 6. 

Of the constants, r and Z are the only ones that require any spe- 
cial methods of measurement. They can be determined approxi- 
mately from a working drawing of the rod itself, but can better 
be determined experimentally as follows : let the connecting rod 
swing as a pendulum on a knife edge, first through the wrist pin 
and then through the crank-pin brasses. Let t w be the. observed 
time of one vibration in the first case, and t c the time in the 
second. Then 

where l w and / c , the lengths of equivalent simple pendulums, become 
immediately known. Now if h w is the distance from the point of 
suspension at the wrist-pin end to the centre of gravity of the rod, 

K (4 - h w ) = q\ 
where q is the radius of gyration about G. Also, 

*.(/.- *.) = $», 

and h w + h c = Z , 

where Z = L + (radius of crank pin) -f (radius of wrist pin). 

Combination of these equations gives 

^ _ Z (Zq — / c ) ^ 

2 z 4, i c 
from which r is obtained by subtracting the radius of the wrist 
pin. The moment of inertia about G is also determined from 

TV 

o 

where W* is the weight of the rod in pounds. 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 233 



In order to study the effects of these forces it will be best to 
follow through an actual case. Curves showing their magnitude, 
direction, and general variation will give a better insight into the 
subject than many pages of formulae. Take for example a small 
6x8 horizontal engine, whose dimensions are as follows : 



W' = weight of connecting rod .... 
W" = weight of piston and cross-head 

/ = throw of crank 

L = length of connecting rod .... 
/ 

n = z 

N — number of revolutions per minute 

a) = number of radians per second . 

r = distance of centre of mass of rod from centre of 
wrist pin ...... 

I = moment of inertia of rod about G 



24 lbs. 
33.6 lbs. 

■333 ft. 
2 ft. 



.1666 



300 

3 I - 62 5 

1. 123 ft. 
4233 



We must now compute the values of 



d 2 a 



for 



d*x , d~y 

w w and IP' 

every io° of the crank orbit. Table No. 1 shows the results 
of these computations. The last column gives the values of 

or the acceleration of the cross-head and piston. This is 



dt 2 

easily obtained from the expression for 

This then becomes 



d*x 
"dt 2 



by putting r = o. 



dhc 
dt 



— /<o 2 cos 6 — 



/n<D 2 (cos 2 6 4- n 2 sin 4 0) 
(i-n 2 sm 2 0)% 



These accelerations are expressed in feet per second per second, 
or in radians per second per second. 

The curves of Plate I show graphically the results of Table 
No. t, and the curves as numbered correspond to the columns 
of the table. The values in the table extend from = o° to 
= 180 , but it is easily seen that for the remainder of the circle 



234 



MECHANICS OF THE STEAM-ENGINE 



columns 2 and 4 will be repeated in reverse order without change 
of sign, while columns 1 and 3 will be repeated in reverse order 
with change of sign. 

TABLE No. 1 

Computation of Accelerations 



e 


d 2 a 


J^x 


d*y 


d*x 


dP 


dp 


dP 


dp 




(1) 


(2) 


(3) 


(4) 


o° 


.OOO 


- 357.746 


.OOO 


~ 388.937 


IO° 


+ 28.I76 


35 I - 2 4i 


- 3 2 -5° 2 


380.589 


20 


55- 6 96 


332.040 


64.016 


356.063 


30° 


81.878 


301.078 


93584 


316.893 


40 


IO5.985 


259.805 


120.310 


265.466 


5o° 


I26.944 


210.192 


143.380 


204.946 _ 


6o° 


I44.847 


154.506 


162.093 


138.913 


7°° 


I58.060 


95- x 94 


175.882 


7L093 


8o° 


166.265 


-34.696 


184.325 


-5.OO9 


90 


I69.O49 


+ 24.715 


187.169 


+ 56.350 


IOO° 


166.265 


81.084 


184.325 


IIO.77I 


IIO° 


I58.060 


132.850 


175.882 


I56.95 1 


I20° 


I44.847 


178.872 


162.093 


I94.465 


130° 


I26.944 


218.388 


143.380 


223.634 


140° 


IO5.985 


250.859 


120.310 


245.298 


150° 


81.878 


276.368 


93-5 8 4 


260.553 


160 


55-696 


294.504 


64.016 


270.481 


170 


+ 28.176 


305-383 


- 3 2 -5° 2 


276.035 


180 


.000 


+ 309.008 


.000 


+ 277.817 



We now come to the calculation of the forces ; namely, Q x , Q y , 
P x , and P y . In the expression for Q y (equation No. 4) the third 
term is merely the upward reaction of the crank pin upon the rod 
due to its own weight. The fifth term is the horizontal thrust 
upon the rod at the wrist pin due to external forces. These two 
terms, being independent of the others, may be considered at any 
time during the discussion. The same applies to the second term 
in the expression for Q x and the second term in the expression 
for P y . For the present we will omit these, or in other words we 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 235 

will first consider the effect of the connecting rod on the crank 
pin due to its own inertia, and call the forces thus exerted on the 
ends of the rod QJ, QJ, PJ ', and PJ. Then 



QJ = M 



, djx 

df 

In d 2 a 



(10) 



n/rl r d~y , ,,-, L — r d~x . , N 

\-M' *- + M' tan a, (n) 

Z cos a df 1 L dt 2 L df 1 v J 






0,'- 



• (12) 

• (13) 



Finally the pressures of the rod on the pins will be equal and 
opposite to those exerted by the pins on the rod, and these latter 
we will denote by (QJ), (QJ), and (PJ). The effect of the first 




Fig. 179 

two of these upon the rotation of the crank can best be studied 
by resolving them along the tangent and normal to the crank 
orbit, as in Fig. 179. Denoting these forces by (T 1 ) and (JV 1 ), 

(T') = (QJ)cosO-(QJ)smO, 

(N') = (QJ) sin + (QJ) cos 0. 

The values of these various forces acting upon the crank pin and 
guides, due to the inertia of the connecting rod alone, are given 
in Table No. 2, and the graphical results in Plates II and 
III. Since the force (T') acts in the same line as the direction 
of motion, the area between Curve No. 3 (Plate No. 2), the axis 



236 



MECHANICS OF THE STEAM-ENGINE 



of X, and any ordinate represents work done upon or by the 
pin, and stored up as kinetic energy in the rod. Since the state 
of the rod as regards kinetic energy is identical at the two dead 
points, we should expect the net area of the curve between O and 
B to be zero, which a planimeter shows to be a fact. The curve 
on Plate III shows the variation of (P y '), or the pressure of the 
wrist pin upon the guides. In this case, however, the curve is 
plotted along the guide itself. All forces are expressed in pounds. 

TABLE No. 2 
Forces acting on Crank Pin and Guides due to Connecting Rod 



e 


(0*') . 


(&/) 


(T) 


(A") 


W) 


o° 


+ 266.649 


.000 


.000 


4- 266.649 


.000 


IO° 


261.800 


+ 16.204 


- 29.503 


260.636 


+ 8.021 


20° 


247.488 


32.329 


54.267 


243.620 


15.386 


3o° 


224.411 


48.229 


70.441 


218.459 


21.528 


40 


193.648 


63.650 


75-7I3 


189.256 


26.024 


5°° 


156.668 


78.144 


69.785 


160.566 


28.725 


6o° 


115. 162 


91.360 


54-053 


136.701 


29.458 


7°° 


70-955 


102.478 


31.624 


120.566 


28.616 


8o° 


+ 25.861 


1 10.895 


-6.21 1 


113.701 


26.493 


9o° 


— 18.422 


115.989 


+ 18.422 


115.989 


23.5 ! 9 


ioo° 


60.437 


117.264 


39.155 


125.687 


20.124 


IIO° 


99.021 


114.434 


53-9IO 


141.400 


16.660 


120° 


133.324 


107-433 


61.746 


159.702 


13.385 


i 3 o° 


162.777 


96.379 


62.743 


178.461 


10.490 


I40° 


186.980 


81.841 


58.028 


195.841 


7.833 


I 5 0° 


205.993 


64.191 


47.406 


210.490 


5-564 


160 


219.511 


44-I5 1 


33-589 


221.374 


3-564 


1 70° 


227.620 


+ 22.489 


+ 17-379 


228.067 


+ 1.736 


180 


- 230.321 


.000 


.000 


+ 2302.31 


.000 



The next point to be considered will be the effect of the external 
force P x . This, as has been said, is due to two causes, the inertia 
of the piston and cross-head, and the varying pressures on the 
cross-head due to steam pressure and friction. We will not con- 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 237 

sider in this case the last causes, but take up the purely theoretical 
case of the crank and reciprocating parts, moving without friction ; 
and neglect the pressures due to the weight of the parts. The 
portion of P x due to this cause we will call P x ". We then have 

PJ' = M»~, .... (14) 
dt 2 V q; 

d 2 x 
where M" is the mass, and — y the acceleration of the piston and 

cross-head. Now, if the component forces acting on the crank 
pin parallel to the axes of AT and Y due to P x " are (Q x ") and 
(Qy")> an d the force at right angles to the guides is (P y "), we will 
have 

(QJ')=PJ', (15) 

(<2/) = ^"tan«, .... (16) 

W) = -(&") (n) 

d 2 x 
The values of —j we have already computed in Table No. 1, and 

these only need to be multiplied by M" to give us P x ". The 
components (Q x ") and ((?/') may also be resolved along the 
tangent and normal to the crank orbit, giving (T") and (W). 
All these forces are given in Table No. 3. Plate IV gives the 
curves of (Q x "), (£,"), (T"), and (N"), while Plate V gives 
the curve of (P y ") laid out along the guide. The curve of 
tangential effort, namely, (T") } must have a net area equal to 
zero, and such is found to be the case. 

Taking now the total inertia effects of connecting rod and 
cross-head and piston, we have the complete curves in Plates VI, 
VII, and VIII. The first of these shows the total horizontal 
and total vertical components of the forces active at the crank pin 
due to inertia. The second shows the total tangential and total 
normal components. A most remarkable resemblance is seen to 
exist between the two curves of tangential effort. In fact, if a 
mass about .535 of the mass of the connecting rod were con- 
centrated in the cross-head, an almost identical curve of total 
tangential effort would be produced by the acceleration of the 
cross-head alone. 



2 3 8 



MECHANICS OF THE STEAM-ENGINE 



TABLE No. 3 

Forces acting on Crank Pin and Guides due to Piston and 
Cross-head alone 



e 


(£*") 


(Qy") 


(T") 


{N") 


W) 


o° 


+ 405.847 


.000 


•OOO 


+ 405.847 


.000 


IO° 


397.127 


-ii.497 


- 80.283 


389.086 


+ n-497 


2O 


371-543 


21.206 


I47.OO4 


341.806 


21.206 


So 


330.671 


27.025 


189.284 


272.543 


27.025 


40 


227.008 


29.848 


200.922 


I93.OI4 


29.848 


5o° 


213-857 


27.522 


181.515 


1 16.381 


27.522 


6o° 


144.952 


21.139 


136.IO2 


54.169 


21.139 


7o° 


74.184 


11.764 


73-734 


14.317 


11.764. 


8o° 


+ 5.227 


-.870 


- 5.298 


.052 


-f 0.870 


90° 


- 58.800 


+ 9-939 


+ 58.800 


9-939 


- 9-939 


IOO° 


115.587 


19.234 


I IO.491 


38.581 


19.234 


IIO° 


I03-775 


25.972 


i45- OI 5 


80.420 


25.972 


120° 


202.920 


29-593 


160.938 


127.088 


29.593 


i 3 o° 


233-357 


30.032 


I59-458 


I73-005 


30.032 


140° 


255-963 


27.581 


143.402 


213.807 


27.581 


i 5 o° 


271.880 


22.738 


116.249 


246.827 


22.738 


i6o° 


282.234 


16.109 


8i.393 


270.723 


16.109 


170 


288.036 


+ 8.339 


+ 41.803 


285.I08 


- 8-339 


180 


- 289.896 


.000 


.000 


+ 289.896 


.000 



Thus far we have neglected the weight of the rod in producing 
pressure on the crank pin and wrist pin, but this can readily be 
taken into account. By referring to equations (4) and (6), we 
see that the effect of this is to add a constant positive quantity 

— M'g to QJ, and a constant positive quantity — — M'g to P y '. 

The effects on the crank pin and guides will be equal and oppo- 

v 

site to these, or we will add constant negative quantities — M'g 

L — r 

and — - — M'g to (<2/) an( ^ (^V)* No change will be produced 

in (<2x')- The values of (Q ¥ ') and (Py), however, change sign in 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 239 

passing the dead points, while the pressure due to the weight of 
the rod is constantly downward. Hence these two values, when 
corrected for the weight of the rod, will be different in the two 
semirevolutions above and below the axis of X. Likewise the 
value of (7*') will be affected to the extent of having the term 

— yM'g cos added. Plate IX shows these various corrected 

curves. The correction to be applied to (PJ) could be shown, 
if required, in Plate III, by a straight line whose ordinates 
would be subtracted from those of Curve No. 3, and in the same 
way the weight of the cross-head might be taken into account in 
Plate V. 

The above analysis is exact and complete, but simpler approxi- 
mate formulae can be deduced from them, which in most cases 
will give results sufficiently close. These approximations will be 
taken up under the special cases to which they apply. 



B. Applications 

(a) The Fly-wheel 

We have seen that the tangential force at the crank pin, or the 
turning effort, is far from constant. In fact, it is necessarily zero 
at the two dead points, and for a single crank reaches a maximum 
somewhere before the 90 position. Hence, in order to secure an 
approximately constant value of the angular velocity, we must 
provide some reservoir of energy in the shape of a large mass 
rotating with the main shaft. This is realized in the fly-wheel. 
Let H'bTeC (Fig. 180) be the curve of tangential forces or turn- 
ing moments of a steam-engine. This is the sum of all such 
forces, due both to steam and inertia. Let T m be the mean 
turning effort found by means of a planimeter. Then 

• Area (b Te) = Area (B'at) + Area ( Cde) . 

The ordinate T m also represents the net resistances to turning due 
to the pull on the belt and to friction reduced to a lever arm 



240 



MECHANICS OF THE STEAM-ENGINE 



equal to the crank throw. Suppose the crank pin to be travelling 
from H' toward C. Then while moving from E/' to B the engine 
will be slowing down, as the tangential force on the crank pin is 
less than the mean tangential force or belt pull. But at B the 
forces are equal, and beyond B the engine is speeding up ; hence 
the minimum velocity of the crank pin is at B. The engine then 
speeds up till E is reached, and hence at E occurs the maximum 
velocity of the crank. Again the engine slows down, and the 
minimum occurs again at B. While the crank is passing from B 
to E, the total work done by the tangential force is proportional 




Fig. 180 



to the area BbTeE. But of this work the belt takes out energy 
proportional to the rectangle BbeE. Hence the area bTe is pro- 
portional to the excess of energy which the engine puts in over 
what the belt takes out, and must be represented as change of 
kinetic energy in the rotating fly-wheel due to change in speed 
between B and E. Call this area AE ft. lbs. 

Let v x and v 2 be the greatest and least velocities of the rim of 
the fly-wheel allowable during one stroke of the engine. Let E x 
and E 2 be the energies stored up in the rim at these velocities. 
Then 



E x = W 



2g 



and 



£i 



E 2 = W V -^ 

2g 



E 2 = AE = ^(v 1 2 

2g 



(very nearly), 



vi). 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 24 1 

If v be the mean velocity of the rim (as given by a speed counter), 

then, since v x and v 2 are very nearly equal, the value of v will be 

given nearly enough by 

« ? 'i + v * 
2 

Also call S the Coefficient of Fluctuation of speed, where 

3 _ i\ — Vi t 

V 

Now factoring the expression for &E we get 

A£ = ( Vl + 7' 2 ) (^ - » 2 )-- - — — - 

If the diagram of turning efforts is drawn accurately to scale, 
&.E can be measured by means of a planimeter, and expressed in 
foot-pounds. Hence 8, the coefficient of fluctuation, can be calcu- 
lated, or if S be assumed, the weight of rim necessary to preserve 
this fluctuation at the given speed of rim becomes known. 




Fig. 181 



If two or more cylinders are working on the same shaft, the 
turning efforts must be combined into a single diagram. Where 
the pistons work on one rod and crank, the phases of the turning 
efforts are identical, and the ordinates of the curves of each cylin- 
der are merely added for each position of the crank, as in Fig. 181. 
Where two pistons work on different cranks at 180 apart, the 
effect is practically the same only that the turning effort of the 

R 



242 



MECHANICS OF THE STEAM-ENGINE 



head end of one cylinder is to be combined with that of the crank 
end of the other. In both of these cases it will be seen that the 
double cylinder presents no superiority so far as speed fluctuation 
is concerned over the single. If, however, the two cranks are at 
90 , the phases are not identical, and the maximum turning effort 
of one cylinder comes very nearly when that of the other is zero. 
The resulting curve of turning efforts is therefore much smoother 




Fig. 182 



than for zero or 180 cranks, and the result will be something 
like Fig. 182. The magnitude of 8 will be greatly reduced, and 
hence a much lighter fly-wheel will be needed. For three cylin- 
ders working at 120 the turning effort is still more uniform. 

Having the curve of total tangential forces, we can construct an 
approximate curve of tangential velocity of the fly-wheel by a pro- 
cess of mechanical integration. From the preceding we have 
approximately 



v 1 = v[i + 



v 9 = v[ I — 



Hence knowing v and 8, the extreme speeds v 1 and v 2 can be cal- 
culated, and laid off at B and E (Fig. 183). Now draw an ordi- 
nate at X at a small distance beyond B, writing the equation of 
energy between B and X, 

2 2 

W^ + (Area xx'd) = ^— • 

2g 2g 



INERTIA EFFECTS- OF THE RECIPROCATING PARTS 243 
From which 



z/ x = y 



vi + 



2,g-(Area xx'fi) 



W 



Similarly at point Y, 



V 2 . 2P-(Area xx'yy') 



and so on. Thus we can obtain the velocity at each point along 
the crank orbit, and a curve of velocities can be drawn in as 
shown along the mean velocity as an axis of X. 




YXB 



Fig. 183 



Having obtained the velocity curve, the curve of space variation 
can be laid in by similar means, as shown in Fig. 184. Divide 
up the velocity curve by means of a number of ordinates spaced 
at equal small distances along I/"C", the semicircumference of 
the fly-wheel rim. Take one of these ordinates at A, the point of 
minimum velocity, v 2 (or at the point of maximum velocity, v^). 
If we consider another equal wheel rotating about the same axis, 



* The areas are, of course, expressed in foot-pounds. 



244 



MECHANICS OF THE STEAM-ENGINE 



and at the mean velocity v, we can study the space variation of 
the actual wheel with reference to axes fixed in this imaginary one.* 
Now consider a point in each of these wheels which when at A 
are coincident. Call the small distance AB, As. The time re- 
quired for the mean velocity wheel to cover this distance will be 



A/ = 



As 




Fig. 184 



d c B X A 



v being constant. But during this same interval of time the actual 
wheel has covered a distance equal to 

As' = v' • At, 

As' 



or 



At = 



v' being the average velocity of the wheel throughout the interval 
As, which will be very nearly xx, the ordinate of the velocity 
curve midway between A and B. Hence, 

As' = - A*, 
v 



* Strictly, we should obtain a new and exact value of v by getting the 
average ordinate of the velocity curve. 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 245 

and the distance that the actual wheel is behind the mean velocity 
wheel at B will be 

s 1 = As'-As = As( V -- 

Choosing now two points which are coincident at B, we can find 
in the same way the distance that the actual wheel falls behind the 
mean velocity wheel during the next interval As or BC. It will 
be 



As" -As = As(—-i\ 



and the total space displacement between A and C will be the 
sum of these, or 

s 2 = As [ — ! — 2 



and so on. These ordinates, s lf s 2 , s 8 , etc., laid off with regard to 
their proper sign, constitute the space variation curve. It will be 
noticed that the actual wheel will be falling behind the mean 
velocity wheel whenever the velocity curve is below the mean axis, 
but will be gaining on it when above. Hence a check on the work 
consists in a closure of the space curve at A again. 

Each step in the above deduction consists of a graphical method 
of forming a curve which is the integral of a previous periodic 
curve. The curve of turning efforts is a periodic acceleration 
curve, and from it a periodic velocity curve is obtained by a pro- 
cess of graphical integration ; and from this a space curve is 
obtained by a similar means. It will be noticed also that the first 
or acceleration curve is quite irregular in outline, or departs in a 
marked way from a simple sine curve. The velocity curve is 
more nearly a pure sine curve, and the space curve still more 
nearly a simple harmonic. Furthermore it can be shown mathe- 
matically that curves formed by successive integrations from a 
complex harmonic, expressed as a Fourier's Series, approach 
indefinitely toward the fundamental harmonic as a limit. Advan- 
tage can be taken of this fact to form an approximate analytic 
expression for the space variation curve, and thus save the great 



246 MECHANICS OF THE STEAM-ENGINE 

amount of labor necessary in the graphical method. To do this 
consider the velocity curve a simple harmonic function of the 
time. Denote the relative velocity of the actual and mean veloc- 
ity wheels by w, where 

w' == v' — V. 

Then by the preceding approximation, 

w = w max sin Kt, 
K being a constant, and z£/ max is a known quantity, being equal to 
v-. The time required for a point on the wheel to cover one 

2 s 

complete period of the curve is — seconds. Hence our equation 
is 

8 . irNt ds 

w = v - sin = — j 

2 15 dt 



v \S i 



irNt ,. 

sin dt. 



15 
and the equation of the space curve is 

8 15 irNt r „ 
s == — v ^ cos \-\C = o. 

2 TvN 15 L 

The maximum values of the space variation will be 

8 15 

s ™*- ±z ' 27rN > 

7V in every case being the number of revolutions per minute. 



27rJV 

If expressed in terms of angular measure, s = aR, and v=R 



60 ' 



where R is the radius of the fly-wheel. Hence, 

8 tvNt 
a = — - cos , 

4 15 

and «max = ± - 

4 

If the angle is measured in degrees, 

<ax = i4°-3 2 4S. 
The labor of constructing the exact curve of turning efforts may 
be greatly abridged by making an approximation to the value of 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 247 

the inertia effect. In fact, the exact value is useless in connec- 
tion with fly-wheel work, as the irregularities and wide variations 
in the steam effect are so great as to utterly mask any small error 
due to this approximation. We may consider, then, that a portion 
of the connecting rod equivalent to one half of its total mass is 
concentrated at the cross-head, and merely compute the tangential 
force due to (M" + \M } ) p x . Furthermore, no great error will be 
committed if we consider p x as a simple harmonic accelera- 
tion, in which case 

p x = — /<o 2 cos 0, 

and since then (T) = (M" + \M<) p x sin 0, 

we will have (T)=-l ^ 2 J /<o 2 sin 2 6. 

The use of these formulae can best be illustrated by applying 
them to a specific example. Consider the curve of turning 
moments of the small 6x8 horizontal engine before mentioned. 
In Plate X are shown the total steam and back pressure lines of 
one forward stroke of the engine, deduced from the actual indi- 
cator card. Resolving the total horizontal force by the graphical 
method into its components at the crank pin, we obtain the curve 
of turning moments due to steam. Adding to this the curve of 
tangential accelerative forces already obtained in Plate VII, or as 
calculated more approximately from the preceding paragraph, we 
get the curve of net turning efforts (Plate XI). These are drawn 
on a scale of 1" = 1000 lbs. The mean turning effort is T m = 692 
lbs., from which A£ = 293 ft. lbs. If the weight of the fly-wheel 
rim be 195 lbs., and its diameter be 30 in., we have, at 300 revo- 
lutions per minute, 

# = 39.270 ft. per sec, 

A£ X 32.2 

d = -r^- ^=.03191, 

195 x(39-27) 2 
or 3.2%. We also have 

; 1 + -J = 39.896, 

and v 2 = v[i ] = 38.646 ft. per sec. 



248 MECHANICS OF THE STEAM-ENGINE 

Now forming the equation of energy at each io° of the fly- 
wheel semicircumference, we get the velocity at each of these 
points, and the final result is shown in Plate XII. 

Again measuring the velocity at the middle of each io° interval, 
and calculating the space variation, we get the space curve, 
also shown on Plate XII. It will be noticed that the mean of the 
two maximum values of s is 

W = .01004 ft., 

and the maximum angular variation will be 

«°max = .45 8 4° 

on each side of the mean. 

Applying the approximate formula based on the supposition 
that the velocity is a simple harmonic function of the time, 
we have 



8 15 1 

S m „ =V v*= 30.27 X .Ol^oq X 



1 TTiV -39 * 27 X - OI 595 x—=. 009972 ft., 

and 14.3248=4576°. 

The difference is seen to be very small, in fact much less than 
experimental errors. 

(b) Counterbalancing 

In the problem of counterbalancing we generally attempt to find 
what mass placed at a fixed position opposite the crank will most 
nearly counteract certain components of the forces acting on the 
shaft in the direction of the crank throw. These forces will be 
due to two causes : ( 1 ) to the unbalanced mass of the crank 
itself, and (2) to normal components of the forces due to the 
inertia of the reciprocating parts. 

We will first ascertain the counterbalance required for the crank 
alone. Since the only force exerted by the unbalanced crank is 
in the direction of the normal to the crank circle, and is constant 
in magnitude, its effect may be completely counterbalanced by a 
mass placed opposite the centre of the crank pin. This, of course, 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 249 

is not possible in all cases, but can be realized in the case of a 
centre crank engine by placing one-half of the counterbalancing 
weight on each crank arm. If M Q is the mass of the unbalanced 
part of the crank, and s is the distance of the centre of mass of 
this part from the centre of rotation ; and if M x is the mass of the 
counterbalance, and s 1 is the distance of its centre of mass from 
the centre of rotation, then 

M s a> 2 = Mis^ 2 , M^M^-, or M x = M % . (1 8) 

when s x is equal to the crank throw. This balance will evidently 
exist for all speeds. 

On considering the effect of the reciprocating parts it will be 
seen from a simple inspection of the variable magnitude of the 
total normal component as exhibited in Curve No. 1 (Plate VII), 
that no fixed counterbalance can even approximately reduce the 
effect to zero. However, it may be of importance to counter- 
balance, if possible, certain components of the forces acting at the 
crank pin. For example, let us see how nearly we can counter- 
balance the total vertical component at the crank pin due to iner- 
tia, viz. y — { Qy), by means of a mass placed opposite to and 
rotating with the crank.* If M 2 is the mass of this counterbalance, 
and s 2 the distance of its centre of mass from the centre of rota- 
tion, its total normal force will be M 2 s 2 <n 2 , and the vertical compo- 
nent of this will be — M^^ya 2 sin 6, which is directly opposed to 
the vertical component of the inertia effect. This can be repre- 
sented by a simple harmonic curve of semi-period equal to it, and 
of amplitude M 2 s 2 w 2 . It will of course reach its maximum value 
at = 90 . As opposed to this is the complex harmonic curve 
No. 2 of Plate VI. Evidently these cannot completely neutral- 
ize one another, but we may find a value of M 2 s 2 which will reduce 
the unbalanced component remaining to a minimum value. The 

* It is readily seen that the vertical component due to steam is, in general, 
continually in one direction, and therefore cannot be balanced ; while the 
horizontal component due to steam balances itself against the cylinder head. 



250 MECHANICS OF THE STEAM-ENGINE 

equation of our complex harmonic curve is (from equations Nos. 
n, 14, and 16) 

j=KG/>+(G") 

Zcosadt 2 Ldt 1 L dt 2 df v yy 

Substitution gives 

— 7 ;zw 2 (i — n 2 ) sin li/r} r 2 9 . ^ 

y = - *-— f — M' — n<D~ sin 

y £{1 -n 2 sin 2 0) 2 L 



±L — t-M'l -/o) 2 cos(9 
L I 

nw 2 (/— fir) (cos 2 + ft 2 sin 4 0) ( — n sin 0) 



(i-/* 2 sin 2 0)2 Vi-/r'sin 2 

r 

+ J/"<! -/co 2 COS0 

/nw 2 (cos 2 + ^ 2 sin 4 0) (— ^ sin 0) 
(i-n 2 sin 2 0)% Vi-^ 2 sin 2 



} 



(20) 



It is evidently impossible to counterbalance, even approximately, 
any terms in the above equation except those which have a semi- 
period equal to ir. Let us examine, then, each term separately, 
rejecting those which have a semi-period less than tr. The first 

term is 

K x sin 

■*- (i-« 2 sin 2 0) 2 ' 

K x being a constant. In this term the denominator is practically 
unity for all values of 0, varying between unity and .9724 when 
n is equal to \. Its effect, therefore, will be insignificant in modify- 
ing the numerator whose semi-period is it. The second term is 

y 2 = K 2 sin 0, 

and this has a semi-period ir. The third term may be broken up 
into three, which can be written 



~ , cos sin . «. ,, cos 2 sin . ~ „, 
^3 = ^-3 1-^-3 7 —T~. 2n \2^~ K z 



sin 5 



Vi-/z 2 sin 2 (i-« 2 sin 2 0) 2 (i-« 2 sin 2 0) 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 25 1 

Of these, the term whose coefficient is X s ' has a semi-period of 

-, since the denominator is practically unity ; that whose coeffi- 
2 

7T 

cient is X s " has a semi-period- ; while the last has a semi-period 

ir. But the value of the coefficient K% u is so small in any prac- 
tical case that we are justified in rejecting all three of the terms. 
The same applies to the fourth term of equation No. 20. Hence 
the only terms which we can expect to counterbalance are 

L {(i-;rsin 2 0) 2 J V ; 

and these must approximate to M^^o 2 sin 6. The best value, then, 
for this counterbalance would be that which gave at its maximum 
ordinate (viz. at 9 = 90 ) a result equal and opposite to that ob- 
tained by putting = 90 in equation No. 21. Hence, 

M 2 s 2 ^ = ^\-^- 2 + M'A. . . (22) 

Or the mass of the counterbalance will be 

M 2 = ^-\-^- + M'r*' 



s 2 L { 1 — tr j 

In our case M 2 is found to be .3438, and its weight is 11.07 lbs., 
when s 2 is taken equal to /. 

The results of using this counterbalance are seen in Plate XIII. 
The difference between the vertical components of the recipro- 
cating parts and of the counterbalance is shown in Curve No. 1. 
This sums up all the more important higher harmonics rejected in 

the above method. It has in general a semi-period -, but by no 

2 

change in the mass of the counterbalance could its effect be appre- 
ciably diminished. This counterbalance will also have an effect 

* Still more approximately, when we neglect I — n 2 , 

Mo=—{f +M>r 2 }. 
s 2 JL 



252 MECHANICS OF THE STEAM-ENGINE 

in diminishing, to a small extent, the horizontal component as 
shown. 

If we wish to counterbalance the horizontal component, a similar 
method must be pursued. Here 

y^(Q x ) = (QJ) + (UJ') = M>^+M"^, . . (23) 

y = A/' { - /V cos 6 - nv? V- nr) (cos 2 ° + ** sin4 0) 1 
I (i-/z 2 sin 2 0)* J 

, i^ff f 72 /i ^ a 2 (cos 2 + 7z 2 sin 4 0) 1 , v 

+ J/" j — /or cos 6 v — '-\- . (24) 

I (i-/z 2 sin 2 0)* J 

In this equation the term 

cos 2 



y\ = K± 



(i-n 2 sm 2 6)* 
has a semi-period -, as has the term 

r sin 4 

ys = x 5 -, 

(i-;* 2 sin 2 0)* 

also. Hence these must be rejected, an 1 the counterbalance, 
whose horizontal component is 

M 3 s S 'd 2 cos 6, 
must approximate to 

(M' + M")fco 2 cosO. 

These agree exactly for all values of and for all speeds when 
M 3 = (M' + M")-> 

In our case M z is found to be 1.7888, and its weight is 57.6 lbs. 
when s% is taken equal to the crank throw /. The effect of this 
counterbalance is to almost entirely neutralize the horizontal com- 
ponent as shown in Curve No. 1 (Plate XIV) but it very much 
overbalances the vertical component Curve No. 2. 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 253 



The problem of counterbalancing, then, is one which cannot be 
given a general solution, but each individual case must be worked 
out to best suit the existing conditions. The case of the loco- 
motive may be cited as one where the balancing of the vertical 
component is important. In certain paddle-wheel ferry-boats with 
single horizontal engines, the unbalanced horizontal component is 
so great as to cause an oscillation of the whole boat backward and 
forward referred to its mean velocity. In some cases the counter- 
balance might be taken to average up the total normal component, 
which gives in our case a weight of 33.95 lbs. at a distance equal 
to the crank throw from the centre. This is very nearly the mean 
of the other two. 

In conclusion will be given some results for a large passenger 
locomotive. The engine examined is one in regular service on 
the road of the Southern Pacific Co., and the following dimensions 
were obtained through the courtesy of the railroad officials at the 
West Oakland shops : 

Type, CW. 



Southern Pacific Locomotive No. 1436 

Total weight of engine . 

Weight of drivers .... 

Diameter of cylinders 

Stroke 

IV' = weight of connecting rod 

IV" = weight of piston and cross-head 

JV'" = weight of side rod 

/ = throw of crank . 

L — length of connecting rod 

/ 

n = — 

L 

D = diameter of drivers 

JV= revolutions per minute, at 60 miles an hour 
w = radians per second, at 60 miles an hour 
r= distance of centre of mass of rod from centre 
of wrist pin .... 

I = moment of inertia of rod about G 



131,400 lbs. 

85,850 lbs. 

20 in. 

24 in. 

549 lbs - 

507 lbs. 

275 lbs. 

1 ft. 

8. 1 1 94 ft. 

.12316 

6.0000 ft. 

280.1 

29-333 

5.0903 ft. 
171. 81 



254 MECHANICS OF THE STEAM-ENGINE 

The last two dimensions were obtained by swinging the rod as 
a pendulum, and it was observed to make 40.50 vibrations per 
minute when suspended from the wrist-pin end, and 43.0 vibrations 
when reversed. 

In the discussion of this case a new member appears ; namely, 
the side rod. The effect of this is easily seen, however. Since its 
centre of mass, which as nearly as could be determined was at the 
middle point of a line connecting the centre of its bearings, travels 
uniformly in a circle relatively to the locomotive, the force neces- 
sary to maintain this path will be directed toward the centre of 
this circle and will be equal to — M"'/uj 2 , where M'" is the 
mass of the side rod. The horizontal component of this force 
will be QJ" = — M/u) 2 cos 0, and the vertical component will be 
QJ" = —MhJ sin 0. The forces active on the two crank pins will 
each be equal to one-half of this, and will be opposite in sign. 
Hence, 

(T'") = °, 
(A-»<) = ^. 

In Table No. 4 are given values of total horizontal force at the 
crank pin of the forward driver, or 

and of the total vertical force, 

(&) = (&') + (ft") + (&'")• 

The curves of Plate XV show graphically the same thing. They 
are of form similar to those obtained for the 6 X 8 engine, but 
the irregularities are even less, this being due to the smaller 
value of 11. They are interesting as showing the enormous magni- 
tude of these forces in a high-speed locomotive. They all increase, 



INERTIA EFFECTS OF THE RECIPROCATING PARTS 255 
TABLE No. 4 







Horizontal 


Unbalanced 




Vertical 


Unbalanced 


e 


((?*) 


Component 


Horizontal 


(&) 


Component 


Vertical 






of Counter 


Component 




of Counter 


Component 


o° 


+ 34666 


-H717 


+ 22949 


O 


O 


O 


IO° 


34047 


"539 


22508 


+ 1574 


-2035 


— 461 


20° 


32084 


IIOIO 


21074 


3170 


4007 


831 


3o° 


29025 


10147 


18878 


4749 


5858 


1 109 


40 


24954 


8976 


15978 


6324 


7531 


1207 


5o° 


20024 


7531 


12493 


7838 


8976 


II38 


6o° 


14572 


5858 


8714 


9244 


IOI47 


903 


7o° 


8770 


4007 


4763 


10460 


IIOIO 


550 


8o° 


+ 2895 


-2035 


+ 760 


1 1408 


"539 


- 131 


90 


-2796 





-2796 


1 201 1 


11717 


+ 294 


IOO° 


8164 


+ 2035 


6126 


12213 


"539 


674 


IIO° 


13045 


4007 


9038 


11977 


IIOIO 


967 


I20° 


17273 


5858 


"415 


11281 


10147 


"34 


I30° 


20975 


753i 


13444 


10158 


8976 


1182 


I 4 0° 


23934 


8976 


14958 


8642 


753i 


mi 


I5O 


26213 


10147 


16066 


6785 


5858 


927 


160 


27829 


IIOIO 


16819 


4664 


4007 


657 


170 


28815 


"539 


17276 


+ 2376 


-2035 


+ 341 


180 


-29117 


+ 11717 


- 17400 












of course, with the square of the speed. The counterbalance neces- 
sary for the reduction of the vertical component to a minimum 
can also be calculated, remembering that the effect of the side 
rod must be taken into account. Hence we have 



M 9 = 



SoL 1 1 — nr J 2 s 2 



and if s 2 is taken equal to the crank throw, M 2 becomes 13.61, 
and its weight is 438.49 lbs. The result of using this counter- 
balance on the forward driver is seen in Plate XV, where the 
unbalanced vertical component is shown in Curve No. 1. This 
unbalanced force reaches a maximum of between 1 100 and 1 200 lbs. 
In addition to this weight must be added another sufficient to 



256 MECHANICS OF THE STEAM-ENGINE 

balance the crank pin, etc. As nearly as could be determined 
by measurement of volumes this weight would amount to 267 lbs., 
or the total counterbalance would be about 705 lbs., if the vertical 
component alone were to be balanced. The weight added to 
balance the crank would, of course, affect in no way the values of 
Curve No. 1 of Plate XIII. The counterbalance for the rear 
driver, when placed at a distance / from the centre of this driver, 
would have a mass equal to merely 

M'" 
2 

or, as nearly as the weight of the crank pin, etc., of the rear driver 
could be determined, the weight of the rear counterbalance would 
be 202 lbs. This counterbalance would exactly neutralize all com- 
ponents on the rear driver. 

3. GOVERNORS 

The governor is intended to regulate variations in speed extend- 
ing over longer intervals of time than can be kept within proper 
limits by the fly-wheel. It may act by opening and closing a 
throttle valve, thus varying the work by varying the initial pressure, 
the cut-off remaining constant. Or it may vary the cut-off, the 
initial pressure remaining constant. 

A. The Fly-ball Governor 

The oldest form of governor is the fly-ball, invented by Huygens, 

and used first on the steam-engine by James Watt. The theory 

of the arrangement is simple. The ball at B (Fig. 185), at the 

extremity of the arm CB, revolves about the axis CD, and is 

hinged at C. When equilibrium is attained, the resultant force 

acting on B must be in the direction CB. Let this force be P. 

Then 

g h 
tan a = ^- 2 = — , 

g. 



4 



GOVERNORS 



25; 



We see that w varies as 



v 



const 



, but h is not a constant, hence 



cannot be constant. The fly-ball governor is therefore static or 
stable. That is to say, if it is rotated at a constant angular 
velocity, it will take 
up a certain definite 
position, and if dis- 
turbed from that po- 
sition, will return to 
it when the disturb- 
ing force is removed. 
It is evident, there- 
fore, that we must 
use some other curve 
than the circle if the 
governor is to be 
astatic or neutral. Let 
DBE (Fig. 186) be 
the required curve. If the ball is to be in equilibrium, we must 
have the resultant force in the direction of the normal to the curve. 




MRu* 



Fig. 185 




tana 



g 



dy, 

dx 



**= 2 4y- 



This is the equa- 
tion of a parabola. 
We also have 



tana=f 
h 



or a) 



•4- 



Hence, if <o is to 
be constant, h 
must be constant, 



258 MECHANICS OF THE STEAM-ENGINE 

which is true in the case of the parabola. The parabolic governor 
is approximated in practice by hinging the arm at the centre of 
curvature C of the working arc of the parabola. This always lies 
on the opposite side of the axis, and therefore the arms must 
cross. For very fine work such governors are much used, as in 
the driving clocks of telescopes. 

A governor always has a certain amount of work to do in open- 
ing the throttle or in moving the valve gear. Suppose the gover- 
nor is running at a certain speed <o, at which it is in equilibrium. 
Now let the speed increase to a value w 1 such that ^ = Km. The 
unbalanced radial force exerted by the weight will be 

MR{^ - a) 2 ) = MRo>\ K 2 - 1 ) = P\ 

For a given ratio of increase in speed it is seen that the force ex- 
erted varies as the square of the speed. Hence the faster the 
governor is run the more sensitive it becomes, as a smaller percen- 
tage of variation of speed will cause it to overcome the frictional 
drag of the moving parts. The increased effect of centrifugal 
force due to this higher speed can be counteracted by a weight 
whose centre of mass lies in the axis, or by a spring. 



B. Shaft Governors 

In many types of slow-speed and in some types of high-speed 
engines the fly-ball governor has been made to actuate an auto- 
matic cut-off. But in most high-speed engines the heavy load 
thrown upon the governor is an objection to its use. A far more 
powerful governor is needed, and this is usually located in the main 
axis of rotation of the engine. The shaft governor is arranged in 
such a way that gravity has no effect upon it, and we substitute 
some other returning force which varies according to the same law 
as that due to the normal acceleration of the rotating parts. This 
can be realized by the use of the helical spring. Before discussing 
the arrangement of such governors we must look into the theory 
of the helical spring. 



GOVERNORS 



259 



Law of the Helical Spring. — Let RQS (Fig. 187) represent 
one- half of a turn of the spring, the sections at R and at S being 
made by a plane 
passing through the 
axis of the helix. 
Let AB be a fixed 
diameter of the sec- 
tion R parallel to 
the axis, and CD 
a similar diameter 
in the section S. 
Now suppose also 
that the line AB is 
fixed in space. Let 
a force P be ap- 
plied at the centre 
S as shown, the line 
CD being free to 
move. The whole 
will now take up 

the position RS\ and CD will take up the position CD'. There 
will be bending and also a certain amount of torsion along 
RS'. Suppose now that CD, instead of being free to move, 
is constrained to remain parallel to its original position. This 
must be the case in the helical spring, for if it were not, CD 
would twist more with respect to one section than with respect to 
the next equal section. This can be realized by applying P at the 
centre or axis of the helix as in Fig. 188. There will now be 
an additional bending moment along S'R equal and opposite to 
the first one, and also an equal amount of torsion in the same 
direction. Hence the bending moments tend to neutralize one 
another, while the torsions add their effects. But the same is true 
of every section of the spring ; so if the spring be long in propor- 
tion to its diameter, there will be no bending at all, but only uni- 
form torsion along the wire. Hence we treat the helical spring as 
a problem in pure torsion. Consider a bar subject to torsion by a 




260 



MECHANICS OF THE STEAM-ENGINE 



force P being applied at the extremity of a lever arm a. If r is 
the radius of the bar, / its length, and the angle through which 
the moment Pa twists it, we have from Strength of Materials, 

Pa = E - — , 

/ 2 

where E is the coefficient of elasticity of the material. Now in the 
helical spring let r x equal the radius of the helix, r the radius of 




Fig. 188 

the wire, and n the number of turns of the helix. If e is the elon- 
gation of the spring for full P, then 

a = r 1} 

since the force acts through the axis of the helix. Also when the 
deflection is small as compared with r lt 

6= e -, 



as then a = — , and 0=2 a. (See Fig. 188.) Finally, 

/= 2 TT1\1I. 



GOVERNORS 26 I 

Substituting these in the equation for torsion, 
Pr x — E 



r x X 2 -nr x n X 2 
Pr* 4 tt/z ?'! 3 4 n 



Hence, e = — ^ — r - =P— — 

Let ^ = O, where C is the ratio of the diameter of the helix 

to that of the wire. 

CV 3 4 /z 



P 



{-. 



Er ) 



C s n j 

So we see that the elongation of the spring varies directly with the 
pull. K is called the constant of the spring. If P is measured in 
pounds and e in inches, K is that number of pounds necessary to 
stretch the spring one inch. 

Forces Active in the Shaft Governor. — Let O (Fig. 189) be 
the axis of rotation of the main shaft about which the wheel rotates 
at an angular velocity w. Let the mass whose centre of mass is at 
G be pivoted to the wheel at Q. Let us consider the forces caus- 
ing moments about Q. 

1. Centrifugal Eorce. — This is the force due to the change in 

direction of the velocity u of the centre of mass. Its magnitude 

will be 

P=MPJ ) 

and its direction will be radial. Its moment about Q will be 

Ph =MRM. 

Llence if Q lies on OG, the moment will reduce to zero. Per- 
manent change in angular velocity or in radius will permanently 
change the magnitude of this force. 

2. Tangential Accelerative Force. — This is due to change in 
the magnitude of the velocity u of the centre of mass, resulting 
from rotation about Q. Its magnitude will be 

P' = M d A 
dt 



262 



MECHANICS OF THE STEAM-ENGINE 



and its direction will be at right angles to OG. The moment 
about <2 will be 



P'y= My 



du 
~di 



Hence if G lies anywhere on a circle drawn on OQ as a diameter, 
the moment will be zero, and if Q lies at O it will have no exist- 
ence. So long as R is constant, it will have no existence, and 
permanent change in to and R will not affect it. 




Fig. 189 

3. Angular Accelerative Moment. — This moment is due to 
change in angular velocity w. The moment will be 



Moment = I, 



dt 2 ' 



Its effect will be felt no matter where Q and G may be. So long 
as co remains constant it will have no existence, hence permanent 
change in o> and R will not affect it. 

Of these three force moments, the first is the most important, 
as being the necessary result of a permanent change in speed. 






GOVERNORS 



263 



We will consider its effect at some length, and calculate its varia- 
tions with considerable care. In this case it will be best to begin 
with the simplest possible arrangement, and proceed to more com- 
plex conditions as the theory is developed. 

Centrifugal Force and Moment. — The simplest or ideal case of 
the shaft governor is shown in Fig. 190. O is the centre of 
rotation, and the weight G moves with its centre of mass on a 
radius. A spring is attached to the centre of mass G, and to a 




Fig. 190 

diametrically opposite point on the rim of the wheel. When the 
whole is rotated about O, the centrifugal force of the weight and 
the pull of the spring are directly opposed. The outward force is 

W 

g 



P 1 = —Ru? % 



and the inward force is 



P* = Ke. 



If these are to be in equilibrium, 
W 



g 



Ru 2 = Ke, 



264 MECHANICS OF THE STEAM-ENGINE 

and if this equilibrium is to be maintained for every value of R, 

W 
while co remains constant, then, since — o> 2 = const, and K= const, 

e must be equal to R, since on no other condition could the equa- 
tion be universally satisfied. Then if R = o, e = o, or if the cen- 
tre of mass G were to move in to O, the tension on the spring 
and the centrifugal force would vanish simultaneously. Such a 
governor would be perfectly astatic, and if the whole were bal- 
anced with an equal and similarly situated weight on the opposite 
side of O, and connected to the first by some form of kinematic 
chain, we would have the ideal case satisfied. It is readily seen 
that in this case 



K 



g 

from which r can be calculated by 

4 C z nK 
r = —E-> 
when E is known. 

In a governor so proportioned, the spring is said to have its 
" full theoretic tension." Practically we can never have this exact 
equilibrium of forces, as the inertia of the heavy masses causes a 
racing or " hunting " action. We must therefore give some stabil- 
ity to the governor by employing less than this full tension. Let 
us call the forces acting outward from the centre of rotation posi- 
tive ( + ), while those which act inward toward the centre are 
negative ( — ). If the weight could move inward till its centre of 
mass coincided with O, the elongation which the spring would 
still retain we shall call e . If there is tension at this point, e will 
be negative, and the force due to <? will be negative. If the ten- 
sion vanishes before we reach this point, and there is compression 
at the centre, then are e Q and the force due to it positive. For 
any position of the weight, 

e=R — <? . 

Now the outward force acting on the weight is 

P 1 =+MRo>\ 



GOVERNORS 



265 



and if we plot a curve between J\ as ordinate, and R as abscissa, 
we will have a straight line through the origin, inclined at an angle 
whose tangent is Mm 2 with the axis of R. The inward force act- 
ing on the weight is 

P 2 =-X(R-e Q ). 

This is also the equation of a straight line, which cuts the /'-axis 
at a distance Ke Q from the origin. Suppose that when R = R lf 
P 1 — — P 2 , and that e = ( + ) . Then our lines will be as shown in 
Fig. 191, where AB = AC. Call P— P x + P 2 , then /> is any un- 
balanced force acting on the weight. The line EAF will show by 



+^0 


+p 

^^%urC i mfi~~ : — ___ 


B 

B 

A D 


4 





-P 


G 

J 


\ F 

?2 



Fig. 191 



its ordinates the magnitude and direction of P. At A, P=o, but 
any positive change or increase in R will cause a negative value of 
P. Likewise a decrease in R will cause a positive value of P. In 
other words, any change in R will cause a value of P which will 
oppose, the change, or the governor is stable. In all engines an 
increase in R must cause a shutting off of steam. Hence, if by 
throwing off load R increases to OD, then in order to maintain 
equilibrium DJ must equal DH, or the speed must increase to a 
value wj, where 

angle HOD = tan M^ 2 . 



266 



MECHANICS OF THE STEAM-ENGINE 



When e is positive, therefore, the engine will run faster on light 
load than on heavy load, but the governor will be stable in every 
position. The amount of change in velocity between no load and 
full load can be made as small as we please by decreasing e Qi but 
by doing so the stability of the governor is sacrificed. If e = o, 
the engine would run at the same speed at all loads, but the gov- 
ernor would be in neutral equilibrium, and of no practical use. 

If, however, e is negative, <? = ( — ), then our lines will be as in 
Fig. 192. Here P 1 AF 1 is the line showing the magnitude and 
direction of P. Any increase in R in this case gives a positive 




Fig. 192 



value of P, and a decrease gives a negative value, or any change 
in R causes a value of P which assists the change. Hence the 
governor is unstable. If for any reason the weight be disturbed 
from its position of equilibrium at A, it will rush to its extreme 
position, say at Z> 2 > The engine must now decrease in speed till 
D 2 H 2 is equal to D 2 J 2 , or till 

angle H 2 OD 2 = tan - x Mw 2 2 . 
The slightest disturbance at this point will cause the weight to 
rush to its extreme inner position at Z> lf and now the speed must 
increase till D 1 H 1 is equal to Z> lm / lt or till 

angle H x OD 1 = tan - 1 M^ 2 , 



GOVERNORS 267 

The engine will fluctuate violently in speed therefore between the 
limits. 



=\/— I 1 -] (minimum), 



0)2 -MV R 2 



and Wl=z \lw \ l ~J?) ( maximum )' 



where R Y and R 2 are the limits of the motion of the weight. 

The value of e must, therefore, always be positive. That is to 
say, the tension on the spring must vanish before G reaches O, or 
the engine must run faster on light than on heavy load.* 

In the governor of Fig. 190 suppose we assume a reasonable 
range of speed fluctuation, say in x at no load, and o> 2 at full load. 
From the valve gear design we know the values of R correspond- 
ing to these loads, or the extreme limits of the weights' motion. 
Let these be R 1 and R 2 . Then we have 

W 

— R 1 u^ = Ke 1 (I) 



~R^i^Ke 2 (II) 



Subtraction gives 

W 

— (Ri^i — i? 2 °v) = -K{?\ — e z) • 

But e 1 —e 2 = R 1 — R 2 . 

Hence, 

W 

— (R 1 a> 1 2 -R 2 a> 2 2 ) = R(R 1 -R 2 ) . . . (Ill) 



and KJ^\^l- 



(IV) 



This gives us the constant of the spring. 

Equation III gives us the relation between the various quantities 
for two definite positions R x and R 2 . Now if we substitute R for 

* Professor John Barr has used a diagram similar to the above. See Sibley, 
Journal of Engineering, 1896. 



268 



MECHANICS OF THE STEAM-ENGINE 



R 2 , and <o for w 2 , where ^ and w are any position and angular 
velocity, we get the general relation between R and o>> 

W 
— (jW - R«?) = K{R X - R) , 

o 



Kg 

W 



<*-*) 



■ (V) 



We have seen that if the governor is to be stable, e must be posi- 
tive, which means that the engine must run faster on light load 
than on full load, or the angular velocity must increase with an 
increase in R. Hence if we plot a curve which shows the rela- 



T 










t 


^ ^ — 


ij 


,. 2 




1 


7 


°2 


u t 







/ i 


l 3 1 


*i 


X 



Fig. 



193 



tion between w (or w 2 ) and R, this curve must constantly ascend 
as we pass to the right of the origin. So long as this is the case, 
the governor is stable, but if in any part the curve falls off as we 
pass to the right, then will the governor race over that portion. 
Equation V gives us the relation between co and R for the ideal 



GOVERNORS 



269 



case of a shaft governor. Taking w 2 as the dependent variable, 
this equation is of the form 

b 
y = a • 



It is the equation of a rectangular hyperbola, having x =0 and 
y = a for its asymptotes (Fig. 193). Hence the curve will always 

ascend as we pass to the right, provided -^ > u^ 2 , which is true 

when <? = ( + ). Care should be taken not to use the curve too 
far to the right, or too near its asymptote y = a, otherwise the 
governor will not have sufficient stability. 

It is of interest to note that if e =o, the hyperbola becomes a 

pair of straight lines, and o> 2 = a = -yp, which is constant. If e is 

negative, we have the hyperbola conjugate to y — a , and this 

will continually fall off as we pass to the right. It is impossible 
to use the negative 
branch of the hyper- 
bola, as the angular 
velocities become 
imaginary. 

The case shown 
in Fig. 190 is not 
adapted for a prac- 
tical design. It is 
difficult to have the 
spring cross the 
shaft, to say noth- 
ing of the interfer- 
ence of two springs 
and weights as soon 
as another is intro- 
duced. A more prac- 
tical arrangement is shown in Fig. 194. If the weights GG do 
not move through too great a range, we may consider their paths 




Fig. 194 



270 MECHANICS OF THE STEAM-ENGINE 

as approximately radii, and work out our design exactly as in the 
preceding case. The result may be close enough in many in- 
stances. The moment about Q due to centrifugal force is 

MRM, 

and the moment due to the returning force of the spring is 

Kea. 

Hence, MR x ^b = Ke l a, (I) 

MR 2 ^ 2 2 b = Ke 2 a. . (II) 

Subtraction gives 

W 

— b (R^ 2 - R 2 w 2 2 ) = Ka (>! - e 2 ) . 

o 

But b fo — e 2 ) = a (R ± — R 2 ) , 

W a 

— b(R 1 <» 1 2 -R 2 a> 2 2 ) = Xa(e 1 -e 2 )-, . . . (Ill) 

o 

^ ^'~y , . . (iv) 

g cr R 1 — R 2 

where R 1 and R 2 , and o^ and w 2 are the extreme values of the ra- 
dius and angular velocity. 

If finally we substitute R for R 2 , and o> for w 2 , these being any 
corresponding values of radius and angular velocity, 

W 

— F\R X ^ - Ra> 2 ) = Ka 2 (R x - R), 

o 



, Kg a* Al \WF-~ mi 

CO — 



IV b 2 R 



(IV) 



And this is the equation of a rectangular hyperbola between <o 2 
and R. 

If close regulation is to be attained, the fact that the weight 



GOVERNORS 



2^1 



does not move on a radius, but on the arc of a circle must be 
taken account of. In Fig. 195 let the centre of mass of the bar 
and weight be at G. Let the spring be attached to the bar at E 
and to the rim of the fly-wheel at A. If the spring is very long, 
as it usually is, we can consider it as always pulling parallel to 
itself. Draw QF perpendicular 
to the direction of the pull of 
the spring, and drop a per- 
pendicular OF on QF. Call 
QG = b, QF = a, QF = s, 
and OF=c. Let QG make 
an angle 8 with QF. Given 
values of c, s, and 8, completely 
fix the position of QG. Now 
draw OG =R and QH = h 
perpendicular to OG. The 
moment about Q due to the 
centrifugal force of the mass 
rotating about O will be 

MRu-h = Moment. 

But — = -, where p is the per- 
p h 




oa 



Fig. 195 



pendicular dropped from O 
upon QG. 

W 

Hence, M^bp = — w 2 b (c cos 8 + s sin 8) = Moment. 



The moment about Q due to the returning force of the spring 

will be 

Kea cos 8 = Moment. 

If these are to be in equilibrium, 



W 

— bu?{c cos 8 + s sin 8) = Kea cos 8. 

o 



W 



\c + s ta.n 8) = Kea. 



272 



MECHANICS OF THE STEAM-ENGINE 



When 8 = $ lt let to = o^ and e — e x . When 8 = S 2 , let w = w 2 and 
<? = <? 2 . The values of S x , 8 2j w i> and w 2 are given by the range of 
action of the valve gear, and the allowable range of speeds. 



W 
g 
W 
S 



biti l 2 (c + stanh l ) = Ke l a, (I) 

ba>£(c + s tSLn8 2 ) = Ke 2 a (II) 




Subtraction gives 



Fig. 196 



Wb 

g 

But 



(otic -\~ (a* s tan 8i — <ti 2 c — o> 2 2 s tan S 2 ) = Ka {e x — e 2 ). 



*i 



e 2 = a sin S x — a sin S 2 



H 1-2 — «* oi " "1 «* 

J^2 f I 

— - <| c((tii—o) 2 2 ) -\-s((tii tan Si— w 2 2 tan 8 2 ) [■ 

= X# 2 (sin Si— sin S 2 ). 

K _Wb\ c(iti 2 - o) 2 2 ) + s (iti 2 tan 8t - o> 2 2 tan 8 2 ) 
£■ ^ ^(sinSi — sinS 2 ) 



(in) 

(IV) 



GOVERNORS 273 

If the governor is arranged as in Fig. 196, so that QF bisects 
the total angle through which the lever acts, then we have 

^ I c (rf _ W2 2) + s tan S L ( Wl 2 + w 2 2 ) 1 = 2 Ka? sin 8 2 , . (Ill') 

and K= m{cW-«i) + s^W + «i)^ 

g { 2^sind J 

The general equation IV gives the constant of the spring which 
will give the assumed speeds at the limits. It remains to be seen 
whether the governor will work correctly at intermediate positions, 
which can be determined by plotting the curve between w 2 and 8. 
Equation III shows the general relation between the various 
quantities for two definite positions 8 X and 8 2 . Now let us write 
this equation so as to show the relation between the definite 
position 8 X and any other position 8. For this latter let the speed 
be o). Then 

— { r(o)! 2 - cu 2 ) + si^ 2 tan Sj - o 2 tan 8) ] = ^ 2 (sin S ± - sin 8). 

Solving this equation for w 2 , we get 

to?(c + s tan SO - -7J7T- (sin 8 X - sin 8) 

co 2 = ^ } 

c-\-s tan 
which may be written, 



J ^-{c + s tan SO - ^ sin 8, \ + ^ sin 8 
c-\- s tan 8 
This equation is of the general form, 

a + b sin # 



(V) 



/= 



<: -+- >** tan x 



The curve is shown in Fig. 197, where a>b and c>s, and also 
in Fig. 198, where a > b and c < s. We must always have a 
portion of the curve where it crosses the axis of Y which ascends 



274 



MECHANICS OF THE STEAM-ENGINE 



as we pass to the right. This can be known by putting x = o in 
the first derivative, and noticing whether the result is positive. 
, _ a + b sin x 



dx 



e -\-s tan x 

b cos x (e + s tan x) — s sec 2 x (a + b sin x) 

{c -\-s tan x) 2 



dx 



be 



= (+)• 



the numerator is positive, or when 



Now c 2 is always positive, hence the fraction will be positive when 

be > sa. 
That the governor will be 
stable all the way from A 
to B (Fig. 197) can only 
be found by plotting the 
whole curve. 

The methods of per- 
manently changing the 
speed of an engine may 
be seen by observing the 
equation of equilibrium 
for the astatic governor 
in one of the simpler 
cases. In Fig. 194 we 

have 

\Ke a 

CD =\ ■ 

^MRb 
Since e is very small in 
most cases, it is very 
nearly true that' 

e__a 
~R~~b 




Fig. 197 



In this case, then, 



w= rv 



a I_ 

Since K is unchangeable in any given spring, we may change the 
speed of the engine by : 



GOVERNORS 



275 



1. Varying -, or by shifting the weight G, or the point of 
b 
attachment of the spring, along the arm. Care must be taken, 
however, that by so doing we do not destroy the relation 

e _ a 
~R~~b 

so e must be changed to meet the new requirements. 




FIG. 198 



2. By changing M the mass, since w = const 



-VM 



The calcu- 



lation for the size of the spring wire, as determined by the formula 

. 8 ChiK 
a = . 



can be considered as an approximation only, as the elasticity of 
steel wire varies greatly with its temper. A sample of the spring 



2/6 



MECHANICS OF THE STEAM-ENGINE 



should be obtained and tested, and if this does not agree with the 
assumed value of K, we can vary the value of a till the proper 
result is obtained. If K is known to start with, we must first 
compute the value of a from 

2 _ Wb f c (0)1 — qj 2 2 ) + s (o)! 2 tan S 1 — <d 2 2 tan 8 2 ) 



K& 



s [ (sin 6\ — sin S 2 ) 

and proceed as before. 

The length L of the spring is determined by the dimensions of 
the governor and wheel, but we must be certain that it is great 
enough to prevent the wire of the spring being twisted beyond its 
elastic limit for the maximum value of e. 




Fig. 199 



Tangential Accelerative Force. — The variations of this force 
cannot, in general, be exactly calculated as in the case of the 

preceding, as the value of the coefficient — depends on too 

dt 

many and too complicated expressions to be put in any useful 
form. The best we can do is to investigate the direction of the 



GOVERNORS 



277 



resulting moment, and find out the relative position of the three 
points O, Q, and G in order that this force moment may not give 
instability to the governor. 

In Fig. 199, let Q be the point of suspension of the arm. 
Draw OG tangent to the path of the centre of mass G, and GQ 
perpendicular to OG. If the weight shifts suddenly from G to 
G ly the wheel meantime rotating counter-clockwise, its velocity will 




Fig. 200 



have to be increased in the direction A G, and hence it will tend to 
hold back, giving rise to a force AC, acting through a lever arm 
A Q which tends to assist the change. In shifting from G to G 2 , 
the moment BD x BQ tends to oppose the change. This effect 
tends to cause the governor to race from G to G 1} but gives it 
stability from G to G 2 . In Fig. 200 the direction of rotation is 
reversed, and the tendency to race is now from G to G 2 , while 
from G to G± greater stability is given to the governor by reason 



278 



MECHANICS OF THE STEAM-ENGINE 



of this inertia effect. In some engines the whole action is from 
G to Gi with the direction of rotation as in Fig. 200. In order to 
check this effect when undesirable, a dash pot may be introduced. 
Angular Accelerative Moment. — The same general remarks 
apply here as in the preceding case, although in the ordinary 
form of governors already discussed the effect is small. In a 
certain special form of governor, however, this angular accelerative 




Fig. 201 



moment produces a most powerful effect. Let MM (Fig. 201) 
be a heavy bar pivoted at Q, and with centre of gravity at G. 
A stud at C is directly connected to the slide valve through the 
rod CB. The position of the governor shown is that of the head- 
end dead point. Now so long as the angular velocity of the 
engine is constant, the governor acts exactly as do those previously 
described, through P causing a moment directly opposed to that of 



GOVERNORS 



2/9 



the spring. But if for any cause the engine slows down, the bar will 
tend to run ahead of the wheel, promptly drawing out the cut-off, and 
increasing the steam supply. Such effect can only occur while the 
speed is changing. Hence the arrangement as shown furnishes a 
powerful means of checking slight variations in speed. The rela- 
tive positions of Q, O, G, and C are evidently such as will cause 
the angular accelerative effect to aid in governing, and Q cannot 
be located at Q', as such a location would cause the angular 
acceleration to destroy the governing. But if C turns about Q', a 
slightly better steam distribution would follow, and hence some- 
times a separate eccentric is used. This is pivoted at Q', while 
the weight is pivoted at Q, and the two are connected by a pin 
of the bar working in a slot of the eccentric. This forms one of 
the latest and most successful of the inertia governors. 



APPENDICES 



APPENDIX I 



At any instant during uniplanar motion, the body and space 
centrodes roll without slipping on one another. * 

It is evident that there is always a point in common, as they are 
both swept up by the 
same point ; but we 
are to prove that they 
are always tangent at 
this point, and that 
equal arcs are swept 
up by the describing 
point in equal times. 

Let Q be a point 
fixed in the body 
(Fig. 202), and let 
the coordinates of Q 
referred to space axes 
be a and (3. Let P 
be any particle of the 

body at a distance r from Q, whose coordinates are x and y. If 
PQ makes an angle with the axis of x, 

x = a + r cos 9 . . . . (1) 
and y = /? 4- r sin $ . . . . (2) 

Differentiating with respect to time, 



r 








s 

B 1 












^/ V^ 






















^""^^ 's' 












%s 















«Ad A 










1 y/ 












\ ^^^ B 


V 


s 


/ 








(3 
1 








<- 


.--c.- 


--*. 


1 




X 











x > 







Fig. 202 



dx 
dt 

dt 



r s'm.0 



dO 



da 
~dt 

^ + rC os0~ 
dt dt 



dt' 
dO 



* Adapted from " Uniplanar Kinematics of Solids and Fluids," by George 
M. Minchin, pp. 79-80. 

283 



284 



APPENDIX 



Now if P is to be the instantaneous centre, its component veloci- 
ties parallel to both X and Y must be zero ; also it is seen that 

— - is the angular velocity of the body in this case. Hence, 
at 



da 
dt 

d$ 
dt 



ro) sin v = o 



4- r<a cos = o 



(3) 
(4) 



Substituting the values of r sin 6, and r cos 6 obtained from (3) and 

(4), in (1) and (2), we get 

d@ 

dt 

x = a > 

a) 

da 

These are the coordinates of the instantaneous centre, and the 
equation of the space centrode can be obtained from them by 
eliminating /. 




Fig. 203 



Now let a set of coordinate axes be fixed in the body with origin 
at Q (Fig. 203.) The coordinates of P referred to these are 77 



APPENDIX I 285 

and y. At the same time that P is the instantaneous centre, let 
these axes be parallel to the space axes, but at any other time, let 
them make an angle <f> with the space axes. In general, 

£ = (x — a) cos <f> + (y — ft) sin <£, 
7) = ( y — (3) cos <f> — (x — a) sin c/>. 

But if a point is on the body centrode, it will at some time be the 
instantaneous centre ; and hence, 

dt 

x — a = , 

CO 

da 



Substitution gives 



o dt 



dp da 

dt jl , dt ■ JL 1 

— cos d> -\ sin <f> = L 

CO CO 

da df$ 

dt dt . 

— cos <p H sin <p = rj. 



These equations express the locus of all points of the body which 
have been or will be the instantaneous centre. This locus or cen- 
trode will be swept up with a certain velocity whose components 
parallel to the axes of £ and 77 will be 

— and _?. where 
dt dt 

[dp] d/3 (da] da 



d£ d \ dt \ , dt . dd> d "\ dt \ . , dt , d<f> 



1 da ) da 

d-n d \ dt \ , dt . ,dd> d 

Tt = dt\^\™*-^ m *Tt+-dt 



Up] d(3 



dt \ . , dt ,dcf> 

— I smc/>H cose/) -77. 

co j co at 



286 



APPENDIX I 



These equations are general all along the body centrode, but at the 
particular time when P is the instantaneous centre, the curves are 
being simultaneously swept up at P, and <f> =o. 



but 




dt 



C da 
dr\ d \ dt 
di~~dt 



da 
~di~"di {' 

{da 
d{3 d \ dj_ 



d \ dt 



-ha 



o> J dt dt 



+ P 



d/3 
dt 



da 



-h a, 



and 



dt n 

y=-+P; 



hence, 0=* *L = * ** = &, 

dt dt' dt dt d$ dx 

or the tangents to the two curves are in the same line, and they 
roll without slipping, since 



Vdx 2 + df = -Vde -j- drf. 



APPENDIX II 



Body 



Space 



Rolling Curves. — The velocity of the point of contact of two 
curves along the curves, when rolled one upon the other, can 
always be expressed in terms of the angular velocity of the moving 
body about the in- 
stantaneous centre, 
and the radii of 
curvature of the 
centrodes. In Fig. 
204 let p be the 
instantaneous cen- 
tre, and let pp x and 
qq x be the infini- 
tesimal elements of 
the space and body 
centrodes respec- 
tively. If we draw 
the common normal 
at p (or q) , and also 
the normals at p^ 
and q x , we will get 
the centres of cur- 
vature A and B, and also the radii of curvature p 1 and p 2 of tne 
centrodes. Let da x be the small angle pAp x and da 2 be the angle 
qBq Y . When the body centrode rolls upon the space till^ coin- 
cides with q 1} the whole system will have rotated through the dif- 

ds 
ferential angle da x — da 2 . But da x = — where pp x 

dO 




Fig. 204 



where 1>i>\ = ds. Also 
Pi 

da 2 — — or d&= dsl ) and -t:=-j-( ]. 

p 2 \ Pl P2 J dt dt\p x p 2 j 

angular velocity of the body about the instantaneous centre, and 

287 



„ de. 

But — is w, the 
dt 



288 



APPENDIX II 



It 
centrode. 



is u, the velocity of the instantaneous centre along the space 

i 



If the curvatures of the 



Hence, finally, -= 

u Pl p * 0) I I 
centrodes are in opposite directions, then evidently — = — 1 

u Pl ?2 

The acceleration of the point in the body corresponding to the 
instantaneous centre can be expressed in terms of u and a>. In 

Fig. 205 p is the moving centre 
which has its position changed 
by the rolling of the curve qq x 
on pp v Let us find the acceler- 
ation of q first in the direction 
of the normal and then in the 
direction of the tangent to the 
space centrode at the instant it 
comes in contact with /. The 
velocity of q will be in a direc- 
tion at right angles to the chord 
qq lf since the body is rotating 
about q 1} and the magnitude of 
this velocity will be v — so) where 
o> is the angular velocity as be- 
fore, and s = qq x . The compo- 
nents of this velocity in the direction of the normal and tangent 
to the space centrode at p are 

v n = S(D cos <f>, 
v t = sot sin <f>. 

dv„ [ ds , dot ] . . d<j> 

-ii« = — <o H s 1 cos d> — sat sin d> — s 

dt J dt dt J ^ v dt' 




Fig. 205 



and 
Hence 



and 



dv, f ds . d<o ) . , , , deb 

— * = J — o) -\ s \ sin <b -f soy cos d> — ^> 

dt {dt dt J r ^ dt 



As q approaches and finally coincides with p, s and <^> become 
equal to zero, or 



dVn 

dt 
dv t 
dt 



ds 
It 



o. 



(O = U(0, 



APPENDIX II 



289 



Hence the point in the body centrode corresponding to the instan- 
taneous centre has an acceleration in the direction of the normal 
to the centrodes only. 

The Instantaneous Centre of Acceleration. — With origin at the 
instantaneous centre, the accelerations of a point of a moving body 
will then be three in number, viz. : — no 2 in the direction of the 

radius vector, r— in the direction at right angles to that line, and 

dt 
uj) in the direction of the normal to the centrodes. Let P (Fig. 206) 
be a point of a body whose centrodes are in contact at O. Take 




Fig. 206 



the origin at O, and let the X-axis coincide with the tangent to 
the centrodes. If we resolve the accelerations parallel to the two 
axes, and denote their sums by X and Y, we get 

v do> . o 

X = — r — sin a — rar cos a, 
dt 

y, dio o . 

= uo)-\- r — cos a — rar sin a, 
at 

or since r • sin a —y, and r • cos a = x, 

X=-y—-X(d 2 , 
at 

F' , doi 

= U(ji + x yuf. 

dt 



290 



APPENDIX II 



If we make X equal to zero, we get the locus of all points which 
have no acceleration parallel to the axis of X, or 



y = 



0) 

dw 
~dt 



This is the equation of a straight line through the origin the tan- 
gent of whose angle with the X-axis is 



tany = — — • 
' do) 



dt 



If we make Y equal to zero, we get the locus of all points having 
no acceleration parallel to the axis of Y } or 



u dt 

y = - + — x. 



This is the equation of a straight line which cuts the X-axis at 

a distance — — from the origin, and cuts the JK-axis at a distance 
dio 

Yt 
u 
- from the origin, hence the angle 8 will be such that 



U<0 

doi 

tan 8 = = — — = tan y, 

u day ' 



dt 



or the two lines cut one another at right angles. The point of 
intersection G of these is called the Centre of Acceleration. The 
axes of X and Y being independent of the direction of r, it follows 
that G has no acceleration at all, or a moving body has a centre 
of no acceleration as well as one of no velocity. 



APPENDIX II 291 

If we call the coordinates of G x a.ndy , these can be deter- 
mined by elimination from the equations of the two straight lines ; 
hence, 

mo — 
dt 

x = 



» 4+ (f 



yo = + 



■•-m 



As an interesting extension of the above, let the accelerations 
of P be resolved in still other ways, viz., in the direction of the 
radius vector and at right angles to it. Denoting these compo- 
nents by A and B, 

A = um sin a — no 2 . 



T> da) , 

B = r \~ uoi cos a, 

dt 



but cos a 

hence, 



X 

? 

A 


sin a 

= Uoi 


y 
~v 

y 


and x 1 

x 2 +y 2 




-.S; 


r 


r 





_ xx 2 -f y 2 da 

B = uu) 



r r dt 

Put A equal to zero, and its equation gives 
2 , & u 

X" -\-)T y = o. 

(0 

The equation of a circle of radius a, which passes through the 
origin, and whose centre lies on the axis of Y, is 



x 2 -j- y 2 — 2ay = o, 
wise if B is zero, we get 



u 

hence our equation represents such a circle where 2a = -. Like- 
to 



2 . o . tilt) 

(I'M 

dt 



292 APPENDIX II 

This represents a circle passing through the origin and with 
centre on the axis of X. This cuts the Jf-axis at a distance 

2 a = — — from the origin. These two circles cut the axes at the 
du) 

It 

same points the straight lines do, and cut each other at the same 
point G, since the angles about G are at right angles. 

As an example of the method of determining the instantaneous 
centre of acceleration, let a cylinder of mass M and radius r roll 
down a plane inclined at an angle a with the horizontal. The 
cylinder is supposed to start from rest, and, after its centre of 
mass has fallen through a distance h, we are required to find the 
coordinates of the centre of acceleration. In this case 



0} 

U 


ii i 

= j Or CD = 

co r r 


u 

r 


From the principle 


of the conservation of energy, 




2 2 




and since 


r 2 

2 




then will 
and 


2/ = 2a/^-, 

* 3 

U 2 

r r^ 


6* 
' 3 


Also since h = s sin a 


(see Fig. 207), 
^ gysin« 
3 




Differentiating, 


du 4 ds 
211 — = 2 or sin a — i 

dt 3 dt 




and as 


ds 
u= — , 

dt 




, dui 
and — = - 
dt 


du 2 

— = -gsma, 

dt 3 

1 du 2 g . 

r = -sina = 

r dt xr 








APPENDIX II 



293 



Then - = — r and — - — = -2;, and we can write down imme- 

0) ClUi 



dt 
diately the coordinates of the centre of acceleration, 

— 2r 2 s 

Xq = 



4r + H 



yo 



4rs* 



4s 2 + r> 




Fig. 207 

In Fig. 207, where r=^ and «$" = 3, these values become 
# = — .0414, 
^ = -.4965. 



APPENDIX III* 



Proof of the General Theorem. — What must be the form of the 
curve xx (Fig. 208), which, during its rotation about a perma- 
nent centre B, will have as a 
normal the line connecting 
its point of intersection P 
with a fixed curve, and a 
fixed point / on that curve? 
In the solution of this prob- 
lem, then, we would consider 
the profile as fixed to and 
carried around by the pitch 
circle. Since, however, it is 
the equation of the profile 
that we are seeking, it will 
prove simpler to exchange 
the relative motions of the 
profile and curve of action, 
thus bringing the former to 
rest, and referring it to a fixed origin. From this point of view the 
curve of action is fixed to and carried around by the pitch circle. 
Let the origin be taken at the centre of the pitch circle. Let 
OC (Fig. 209) make an angle \f/ with the axis of X, where OC is 
a radius vector moving with the curve of action. The equation 
of the given curve of action may then be expressed in terms of 
the parameter «//, or its equation will be 

F(x,y,^)=0 (1) 




Fig. 208 



* For the following beautiful method of deducing the equation of the 
gear-tooth profile from that of its curve of action, the author is largely 
indebted to Mr. A. V. Saph of the University of California. 

294 



APPENDIX III 



295 



Let the equation of the required profile be 

x = 4>W), y=f(+) (2) 

At the point P where the two curves intersect, both relations must 

be satisfied, or 

F^if/), /(if/), y\ = o . . . . (3) 

The slope of the tangent to the profile will be 



dx <fi'(if/) 



(4) 




Fig. 209 

If the radius of the pitch circle is a, and the constant angle COI 
is a, the slope of the chord PI is 



a sin (if/ — a) — /(if/) 
a cos (if/ — a) — cf>(if/) 

Since these are to be at right angles, 

/'(if/) a sin (if/ - a) - /(if/) ^ ^ 
cf>'(if/) a cos (\p — a) — cf>(ij/) ' 

/(if/) _ a cos (if/ — a) — <f>(if/) 
^j~ ~ a sin (if/ ~ a)-/(if/) 

/'0/0 = ~p\ a cos («A - «) - <K«A) 
<f>'(if/) = P lasm(if/-a)-/(if/)l 



(5) 



or 

Let 
and 



(6) 

(7) 
(8) 



296 



APPENDIX III 



where p 
equation 
from (7) 
(7) and 
As an 
action a 
this case 
hence, a 



may be a constant or a function of if/. Differentiating 
(3), and substituting in it the values of /'(«/>) and <f>'(\j/) 
and (8), we can obtain an expression for p, after which 
(8) may be integrated to give the required equations, 
example of the above method let us take as a curve of 
circle of radius b tangent to the pitch circle at I* In 
we will take OC (Fig. 209) as coinciding with 01; 
= o. (See Fig. 210.) 




Fig. 210 

The equation of the curve of action is 
x 2 +y 2 -2x(a+?>) cosil/—2y(a+d)smij/-{-(a+Z>y—P=o (1) 
and the equations of the profiles are 

x = <j>W), y=/W) . ..." (2) 

At the point of intersection of the two curves at P 
</>0A) 2 +/GA) 2 - 2{a + b) cos ^ • 4>(#) 

-2(^ + ^)sin^./W + (^ + ^) 2 -^ 2 = • (3) 



* It is to be observed that there is no rolling between the circles. 



APPENDIX III 297 

The slope of the tangent to the profile is 
dy /'(if/) 



dx cj>'(}f/) 

The slope of the chord PI is 

a sin if/ — /(if/) 
a cos if/ — 4>($) 

Hence, /W = _ a cos ^ - <j>(if/) 

<f>'(if/) a sin if/ — /(if/) 



(4) 



(5) 



(6) 



or /'(if/) = -p\a cos if/ -<j>(if/)\ ... (7) 

and <t>'(if/)=p{asmif/-/(if/)\ . . . (8) 

Differentiating equation (3), and rejecting the factor 2, we 
have 

/WO -/W + <K<A) • 4> '« + (<*+<*) sin ^ ' *ty) 
— (0 + £) cos ^ • c/>'(^) — (0 4- <*) cos if/ -/(if/) — (a + b) sin if/ -/'(if/) = o, 

and substituting for ^>'(^) and/'(^) their values from (7) and (8) 
we obtain 

— /(«A)p * * cos i// + p • /(i/r) • <f>(if/) + <f>(if/)pa sin «/> 

— p • <f>(\j/)/(\J/) + (a+b)smif/ - cf>(if/) — (a+b) cos if/- pa smif/ 

+ (a+b)cosif/.p -/(if/) -(a + b) cos ^ ./(^) 

+ (a + b) a smxp p cosi\/— (a+b) smip- p' ^>(\p) = o . . (9) 

_ (a + b) cos i/r • /(if/) — (a + b) sin ^ • <f>(if/) 

p ~ b cos i/f • /(if/) — b sin if/ - cf> (if/) 

a + b 
= — ! — = constant. 
b 

Differentiating equation (8), we have 

cf>"(if/) = p(a cos if/ — /'($)) = p a cos if/ — p 2 a cos ^ - p 2 cf> (if/) 

<j>"(if/)+p 2 <l>(if/) = pa(i+p)cosi!/ (10) 



298 



APPENDIX III 



A particular solution of this may be seen by inspection to be of 
the form A' cos ij/ whose second derivative is —AT cos i{/. Then 

— K cos \p + p 2 K cos if/ = pa ( i + p) cos \j/, 

x== pa(i+p) =J >a_ == 
p 2 - i p - i 

The particular solution, then, is 

<£(^) = (a-\- b) cos «/f. 
The solutions to <£"(«/0 4 p 2 (cj> (}{/)) = o are 

and <£ (»/>) = Atf-M. 

Then <£ (^) = (a + £) cos ^ 4- ^'p* + Atf-'W . (i i) 

When ^ = o, <f>(if/) = a, and ^ -f- ^4 2 = — ^, 

- _ H±i tf cos ^ + ^ + ^ 2 cos ^ + ^^ Mi*** + ^-**,} 






/'(,/,) = (* + ^) cos^ + ^t-^M^P'A-'r^-^S, 



/(,/,) = (> + £)sin^ + 



a-\-b { A x e { rt , A 2 e-^ 



(12) 



£ 1 fp — *]p 

When if/ = o, /(^) = o, then ^ — ^ 2 = °> but ^ + A 2 = — £ ; 

hence, A 1 = A 2 — 

2 

Then, finally, 

x=<j>(*l/) = (a + 6) cosxp—b <j |- =(a+^) cos^— b cos pij/, 

I 2 J 



y=f(^j)=z{a-\-b) sm\\/—b 



e i(>ty — e -ipty ) 



21 



j. = (#4<£) sin^— <5 sinpi/f, 



# = (# + £) cos if; — b cos a if/, 

b 

y=(a-\-b) sinxf/ — b sin g "J" if/, 
which are the equations of the epicycloid. 



APPENDIX III 299 

If there is a second pitch circle working within the first at the 
pitch point I, and whose profile works on the same circular curve 
of action, we could deduce the equation of its profile by consider- 
ing the circle of radius b to touch in internal contact. In this 
case the sign of b would become negative, or 

x = (a — b) cos if/ + b cos ~ if/, 

— b 

y = (a — b) sin if/ -f- & sin if/, 

— b 

which are x = (a — b) cos if/ 4- b cos ~ . if/, 

b 

y— (a — b) sin if/ — b sin ~ if/, 

b 

the equations of the hypocycloid. 

In the case of the rack we must make a equal to infinity. 

It is therefore necessary to move our origin from O to Q, and 

express our equations in terms of the parameter (see Fig. 210) 

instead of if/. In this case 

y=y, 

x' = x — a, 

(from the property of the epicycloid). Substituting these values 
in the equations of the epicycloid, 

x' = (a + b) cos - - b cos £±i $ - a , 
a b 

y' = (a + b) sin -6-b sin ^±^ 0. 
a b 

Putting a = 00 , and evaluating the indeterminate forms, these 

become 

x' = b(i — cos0), 

y' = 6(0- sin 0). 
These are the equations q( the cycloid. 



300 



APPENDIX III 



Another important case is that in which the curve of action is a 
straight line through I and making a constant angle with 01. In 
Fig. 211 let PIC be the straight line of action, with / as the pitch 




Fig. 2ii 



point. In this case we take the radius vector OC as the perpen- 
dicular let fall upon the curve of action from the origin, a is 
constant, and OC =p = constant. The equation of the curve of 

action is 

x cos \j/ +y sin \p = a cos a =p . 



while those of the profile are 

* = <M), y=A*l>) • • • 

At the point P 

<f> («/r) cos \p +/(«/') sin i}/ =p 

The slope of the tangent to the profile is 

dx $\f> . 

and that of the line PI is 

a sin (if/ — a) — /(«ft) 
a cos (if/ — a) — <£(«//) 
Expanding equation (5), 

a cos a sin if; — a sin a cos \p — f(\J/) _ p sin \\i — q cos ij/ — f(ifi) 
a cos a cos \J/ -j- a sin a sin ^ — </>(^) / cos ^ + q sin ^ — <f>(\fi) 



(^) 
(3) 

(4) 
(5) 



APPENDIX III 301 

where p and q are constants having the values shown in Fig. 211. 
Hence, 

/W /sin^-gcos^-/M = 

<f>'(if/) ' /cos \p + q sin \p -cf>(ij/) ' ' \) 

or / r (^)=- P 5/cos^ + ^sin^-^)J . . (7) 

*'ty)=p{/sin^-7cos^-/ty)} . . (8) 
Differentiating the equation of the curve of action (3), 

<jE)'(^) cos if/ — <£(i/>) sin if/ +/(if/) sin 1// —/(if/) cos ^ = o, 
and substituting from equations (7) and (8), 
p J/ sin ^r — q cos 1// —/(if/) \ cos if/ — cf>(if/) sin ^ 

— p \p cos ^ + q sin ^ — <£(^) j sin ^ —/(iff) cos if/ = o, 

from which we find 

<£(t//) sin i/f —/(if/) cos «/r 
^ ~~ <£(^) sin if/ — /(«//) cos i/> — / 

which is a function of i/^. We now have from equation (7) 

■^lVmt7> m *^^^-^ (9) 

and from equation (1) of the curve of action, 

()= _^_yw^ . . . (I0) 

7 COS if/ cos ^ ' 

Substituting (10) in (9), 

/'(if/) COS if/ 

= -/sin^+/W , 2 sin ^ + sin ^ cos ^ +/ sin ^ j 
psmifz—f^—qcosif/ 1 r r * r r y vry> 

_ (/ sin ^ —/(if/) — ^ cos i/^) (/ sin 2 t^ —/(if) sin t//) 
sin »// —/(if/) — q cos ^ 

. /'(if/) cos if/ -[-/(if/) sin if/ =p sin 2 if/ . . . (n) 

In the same manner we may show that 

— <f>'(if/) sin if/ + cf>(if/) cos ^ =/ cos 2 if/ . . (12) 



302 



APPENDIX III 



These are the differential equations of the profile. They may be 
written 

sin 2 ^ 



/W+/Wtan^=/ 



and 



<£'0/0 -<K</0 cot </, = -/ 



COS if/ 
COS 2 \j/ 



sinxj/ 



and are of the form y' + Py = Q. 

The integrating factor for the first will be 



and for the second, 



^-Jcot^ d>p _ ^-log sin^ _ cgc ^ # 



The solutions are 



f(xf) sec xj, = pj ^^ sec $d$=pj tan 2 f dij,=p(tm $-if) +A U 

</>({}/) CSC if/ = 

/~COS 2 \b C 

—p\ — : — ^ esc \}/di}/= — p\ COt 2 lj/ d\j/= — p(—COt\f/ — lf/) + A 2 . 

When ^ = o, /(»//) = o, <£(^) =/. Hence, A x = o directly, and 



^sin^r r T J ^y sin^ J \i 



>in 2 ii^ _ \ 



2 sin 
T 

2 



which last expression becomes zero when xp = o; hence, A 2 = o. 

Hence, finally, x = <f>(\fi) = p (cos \p -\- if; sin \p), 

y = f(if/~) =p (sinij/ — i}/ cos if), 

which are the equations of the involute of a circle whose radius is/. 
When the radius of the pitch circle becomes infinite, we have 
the special case of a rack. But when a — oo, p — a cos a — oo also. 
The radius of curvature of the involute for any value of if/ is q=p$, 
which is infinite when p is infinite for all values of ij/ except if/=o. 



APPENDIX III 



303 



Hence any finite portion of the rack tooth is a straight line. The 
slope of the tangent to the profile is 

dy 

dy_<fy_ 
dx ~ dx ^' 

dj> 

or the tangent to the profile always cuts the axis of x at an angle if/. 
At the point S (Fig. 212), where the involute cuts the pitch circle, 

x*+f=a\ 



■Ac 



4vtf 2 -/=-^= tana - 
p p 




Fig. 212 



Hence the tangent at -S cuts the axis of x at an angle equal to 
(tan «). The angle SOX = if/ — a = tan a — a. Hence the invo- 
lute cuts- the radius c^S at an angle 8, where 

8 = tan a — (tan a — a) = a. 

In the case of the involute rack, therefore, the teeth are straight 
lines inclined at an angle a with the normals to the pitch curve. 






M 



INDEX 



Acceleration of connecting rod, angular, 
176. 
curve of piston, 174-175. 
of crank pin, 174, 176. 
of points in piston-crank chain, 173- 

180. 
of slide valve, 219. 
of wrist pin, 174, 177-178. 
Accelerations, table of, for steam-engine 

parts, 234. 
Accelerative force moment, angular, 262, 
276-278. 
tangential, 261. 
Action, angles of, 72, 84-85. 
Admission position of crank, 187. 
Advance of eccentric, angular, 183. 
Angles of action, approach, and recess, 
72, 82-84, 92-93, 98-100. 
of maximum efficiency of spiral gears, 
124-13 1. 
Angular acceleration of connecting rod, 
176. 
accelerative force moment, 262, 276- 

278. 
advance of eccentric, 183. 
velocity of connecting rod, 171-173. 
determination of, by instantaneous 

centre, 17. 
of universal joint, 27-28. 
ratio in gearing, 67-68. 
Annular gears, 76, 96. 
Appendices, 283-303. 
Approach, angle of, 72, 82-84, 9 2- 93> 

98-100. 
Astatic governors, 257, 264, 269. 
Axis of rotation, 4. 
Axle, 22. 



B 



Base circles, 
cones, 115. 



Beale's gears, 141. 
Bearing, 9, 21. 

Belts between crossing and intersecting 
shafts, 62-63. 

between parallel shafts, 36-55. 

diagram for, 41. 

effects of centrifugal force on, 39-41. 

length of, for stepped cones, 43, 44, 

47, 49- 
losses in, 42. 
stresses in and power transmitted by, 

37-43- 
transmission of rotation by, 36-55, 
62, 63. 
Bevel gears, 114. 
method of cutting, 117. 
method of draughting, 116. 
Body centrode, 5, 285. 

construction of, 6. 
Bridge of valve ports, design of, 189. 



Cams, cylindrical, 158-159. 

disk, 155-158. 
Central position of valve travel, 182. 
Centre of acceleration, 289-293. 
instantaneous, 5, 33, 68, 171, 177, 284, 
289. 
Centrifugal force moment in shaft gov 

ernors, 261, 263-276. 
Centrodes of motion, 5 A 32, 36, 283. 
of steam-engine connecting rod, 7. 
rolling of, 6, 283-293. 
Circular pitch, 71. 

normal of spiral gears, 120. 
Coefficient of fluctuation of fly-wheeJ 

velocity, 241. 
Combination of elementary forms of 

plane motion, 4. 
Compression position of crank, 187. 
of piston, 220. 



305 



306 



INDEX 



Conchoid, equation of, 148. 

radius of curvature of, at X-axis, 148. 
Conchoidal straight line motions, 146- 

148. 
Cones, describing, 114. 

pitch, 114. 

stepped, 43-54. 
Conjugate method of forming tooth out- 
lines, 95. 

profiles, 69. 
Connecting rod of steam-engine — 

acceleration of points of, 173-180. 

angular acceleration of, 176, 179. 

angular velocity of, 172-173. 

centrodes of motion of, 7. 

forces due to inertia of, 235-236. 

instantaneous centre of, 12, 171, 177. 

position of points of, 165. 

velocity of points of, 167-168. 
Constrainment of motion, 8. 
Counterbalancing of crank, 249. 

of locomotive parts, 253-256. 

of reciprocating parts of steam-engine, 
249-253. 
Crank pin, acceleration of, 174, 179- 
180. 

forces active at, 228-239. 

position of, 165. 

velocity of, 168. 
Creeping of belts, 42. 
Critical angles of spiral gears, 124-131. 
Crossed belts, in stepped cones, 43-44. 

links, 208. 
Cross-head, forces due to inertia of piston 

and, 236-237. 
Crossing shafts, transmission of rotation 

between, 31, 35, 65, 117, 141. 
Crown wheel, 115. 
Curves of action, 70, 71, 294-303. 

pitch, of gears, 67. 
of cams, 155-157. 
Cut-off position of crank, 187. 

of piston, 220. 
Cycloidal straight line motions, 142-146. 

system of gearing, 73-85, 296-299. 
Cylinder of steam-engine, back pressure 
in, 224. 

clearance in, 224. 

forces due to steam in, 223-229. 

power developed in, 226-227. 
Cylindrical cams, 158-159. 



Describing circle, 75. 

standard size of, 78. 
Diametral pitch, 72. 

normal of spiral gears, 121. 
Dimensions of teeth (standard), 71, 79-80. 
Disk cams, 155-158. 

curves of, for constant acceleration, 
156. 

limitations of action of, 158. 
Double universal joint, 30. 
Dynamics of the steam-engine, 222-279. 

forces due to steam pressure, 222-229. 

governors, 256-279. 

inertia effects of the reciprocating 
parts, 229-256. 



Eccentric of steam-engine, 181. 

angular advance of, 183. 

throw of, 183. 
Ellipse, 106-113, 144-145. 

method of calculating perimeter of, 
109. 

method of drawing, 112. 
Elliptic functions, complete (table of), 
108. 

gears, 106-114. 
Engine, steam. (See Steam-engine.) 
Epicycloid, 74-75, 94~97. M3. 2 9 8 - 

spherical, 114. 
Exhaust port of steam-engine, design of, 
189. 

position of crank, 187. 

position of piston, 220. 



Fluctuation of fly-wheel velocity, coeffi- 
cient of, 241. 
Fly-ball governors, 256-258. 

circular, 256. 

parabolic, 257. 
Fly-wheel, analysis of, 239-248. 

coefficient of fluctuation of velocity 
of, 241. 

space variation, curve of, 243-248. 

velocity curve of, 242-243. 
Forces active at crank pin, 228. 

at wrist pin, 227. 

in shaft governor, 261, 268. 






M 



INDEX 



307 



Fourier's Series, application of, to fly- 
wheel analysis, 245-246. 

Friction gearing, 32. 

Functions, complete elliptic (table of), 
108. 

G 

Gearing, friction, 32. 
toothed, 66-141. (See Toothed Gear- 
ing.) 
Generatrix, 114, 115, 138, 140-141. 
Geometrical progression of speeds in 

stepped cones, 51-53. 
Gooch link, 208. 
Governors, 256-279. 

angular accelerative force moment in, 

278-279. 
astatic, 257, 264. 
centrifugal force moment in, 263- 

276. 
changing speed of, 274-276. 
fly-ball, 256-258. 
inertia, 278-279. 
parabolic, 257. 
regulation curves of, 268-269, 270, 273- 

275- 

sensitiveness of, 258. 

shaft, 258-279. (See Shaft Governors.) 

static or stable, 257, 265. 

tangential accelerative force moment 
in, 276-278. 

unstable, 266-267. 
Grant's odontographic tables, 82, 90. 

tables for worm tools, 136-137. 
Gridiron valve, 208-211. 

double admission in, 211. 
Grooved friction wheels, 34-35. 
Guides, forces acting on, due to inertia 
of parts, 236. 

forces acting on, due to steam, 227. 

prismatic, 142. 

H 

Hart's straight line motion, 153-154. 

Helical spring, laws of, 259-261. 

Higher pairs, 8. 

Hobs for worm gears, 136. 

Horizontal transmission by wire ropes, 

55-58. 
Hyperbola, rectangular, 224, 268, 270. 
Hyperboloidal gears, 137-141. 



Beale's, 141. 

failure of, analogy of, to spur and 

bevel, 140. 
involute spiraloid teeth of, 140- 141. 
relation between radii and angles of, 
I37-I39- 
Hypocycloid, 6, 74, 75, 78, 143, 299. 
spherical, 114. 

I 

Inclined transmission of power by wire 

ropes, 58-62. 
Inertia effects of reciprocating parts, 
229-256. 
application of, to counterbalancing, 

248-256. 
application of, to fly-wheel analysis, 

239-248. 
moment of, experimental determina- 
tion of, 232. 
Inertia governors, 278-279. 
Inside gearing, 76, 96. 

lap of valve, 182. 
Instantaneous centre, 5, 33, 68, 171, 177, 
284, 289. 
application of, to machines, 7-17. 
determination of, in machines, 10. 
location in threes on straight line, n. 
of acceleration, 289-293. 
Interchangeable system in gears, 77-78. 
Interference in gearing, 88, 97. 
Intersecting shafts, transmission of ro- 
tation between, 25-31, 35, 64, 
114-117. 
Introduction, 3-18. 
Inverse curves, property of, 150. 
Inversors, 150-154. 
Involute of circle, 46, 85. 
spherical, 114-116. 
spiraloid, 141. 
system of gearing, 85-93, 300-303. 



Journal, 9, 21. 
design of, 21. 



J 



K 



Kinds of motion, 3-4. 

of uniplanar motion, 4-5. 
Kinematics, definition of, 3. 

of the steam-engine, 163-221, 



3o8 



INDEX 



Laps of valve, 182. 
Lead of valve, 183. 
Lemniscate straight line motions, 149- 

150. 
Lemniscoid, 149. 
Limacon, 96. 
Link, Gooch, 208. 

Stephenson, 198-208. 
Locomotive, counterbalancing of, 253- 

256. 
Lower pairs, 8. 

M 

Machine, definition of, 8. 
Machinery of transmission, 21-159. 
Mean pressure, total, 226. 

turning effort, 229. 
Mechanics of the steam-engine, 163-279. 
Meyer valve, 211-215. 

auxiliary circle for, 214. 

diagram for, 215. 
Mitre wheels, 116. 

Moment, angular accelerative force, 262, 
276-278. 

centrifugal force, 261, 263-276. 

of inertia, experimental determination 
of, 232. 

tangential, accelerative force, 261. 
Motion, elementary machine, 8. 

helical screw, 3. 

kinds of, 3. 

relative, 9. 

spheric, 3. 

uniplanar, 4. 
Motions, straight line, 142-154. (See 

Straight Line Motions.) 
Multiple-cylinder engine, forces due to 
steam in cylinders of, 226. 

N 

Net horizontal steam effort, 225-226. 
Non-circular spur gears, 105-113. 
Normal component of forces at crank 
pin, 228-235. 
component of forces at wrist pin, 227. 

O 

Obliquity of involute teeth, 89. 
Oldham's coupling, 24. 
Open link, 208. 
Osculating spur gear, 124. 



Outside lap of valve, 182. 
Overtravel of valve, 189. 



Pairs of bodies or elements, 8. 

higher and lower, 8. 
Palmer's graphical solution for stepped 

cones, 47-49, 53. 
Parabolic governor, 257. 
Parallel bars or cranks, 23. 
motions, 142-154. (See Straight Line 

Motions.) 
shafts, transmission of rotation be- 
tween, 23, 24, 32-35, 36-64, 66-113. 
Peaucellier's straight line motion, 151- 

153- 

Pin-tooth system of gearing, 94-100. 

interference in, 97. 
Piston, forces due to inertia of cross- 
head and, 236-237. 
forces due to steam pressure on, 222- 
229. 
Piston-crank chain of steam-engine, 164- 
180. 
accelerations of points of, 173-180. 
relative positions of points of, 164-167. 
velocities of points of, 167-173. 
Pitch curves, 33, 67, 295-303. 

surfaces, 66, 114, 137. 
Pivots, 23. 
Plain slide valve, 181. 

variable cut-off with, 195-208. 
Position, central, of valve travel, 182. 
curve of piston of steam-engine, 167. 
curve of slide valve, 184. 
of valve relatively to crank, 180-219. 
of valve relatively to piston, 219-221. 
Power developed in engine cylinder, 

226-227. 
Prismatic guides, 142, 
Profiles, 67, 294-303. 

Pure rotation, transmission of, by means 
of belts, 36-54, 62-65. 
transmission of, by means of friction 

wheels, 32-35. 
transmission of, by means of ropes, 

55-62. 
transmission of, by means of toothed 

wheels, 66-141. 
transmission of, through a rigid inter- 
mediate member, 21-31. 



INDEX 



309 



R 

Racing of governors, 266-267, 2 74- 

Racks, 74, 76, 86, 299, 302-303. 

Radius of curvature of Stephenson link, 

203-205. 
Rectilinear translation, 4, 142-154. 
Reference plane for uniplanar motion, 4. 

position of body centrode, 5, 6. 
Regulation curves of shaft governors, 

268-270, 273-275. 
Relative motion, 9. 
Release position of crank, 187. 

position of piston, 220. 
Reuleaux's graphical solution of the 

stepped cone, 45-47. 
Rigid bodies, definition of, 3. 

couplings, 21. 
Robert's straight line motion, 145. 
Rolling curves, 287-293. 
normal acceleration of contact point 
of, 288. 
Rolling of centrodes, 6, 283-286. 
Rope transmission of power, 55-62. 

(See Wire Rope.) 
Rotation, 4. 
transmission of, between crossing 
shafts, 31, 35, 65, 117-141. 
between intersecting shafts, 25-31, 

35. 64. ii4-"7. 
between parallel shafts, 23, 24, 31, 

32-35. 3 6 - 6 4. 66-113. 
between shafts in line, 21, 32. 
by means of belts and ropes, 36-65. 
by means of friction gearing, 32-36. 
by means of toothed gearing, 66- 

"3- 
through a rigid intermediate member, 
21-32. 



Shaft governors, 258-279. 

changing speed of, 274-276. 

forces active in, 261-263. 

ideal case of, 263-269. 

inertia, Z78-279. 

regulation curves of, 269, 270, 273-275. 
Shafting, 22. 

Sheaves for wire ropes, 62. 
Side rod of locomotive, counterbalance 
for, 254. 



Simple engine, forces due to steam in 

cylinder of, 223-226. 
Skew gears, 1 17-14 1. 
Slide valve, 180-221. 

acceleration curve of, 219. 

gridiron, 208-211. 

laps of, 182. 

lead of, 183. 

Meyer, 211-215. 

plain, 180-208. 

position curve of, 184. 

problems in connection with, 190-195. 

variable cut-off in, 195-208, 215, 217. 

velocity curve of, 218. 
Space centrode, 5, 284. 

variation curve of fly-wheel, 243-248. 
Speed of shaft governor, method of 

changing, 274-276. 
Spheric motion, 3. 
Spherical epi- and hypo-cycloid, 114. 

involute, 114. 
Spiral gears, 117-135. 

circular pitch of, 103, 120. 

critical angles of, 124-130. 

diagram for laying out, 131. 

diametral pitch of rotary cutter for, 121. 

formulas for calculating angles of, 
121-124. 

osculating spur gear of, 124. 

proof of action of, 117-120. 

selection of cutters for, 123-124. 
Spring, laws of helical, 259-261. 
Spur gearing, 66-113. 

circular, 66-105. 

non-circular, 105-113. 
Static governors, 257, 265. 
Steam-engine, acceleration of parts of, 
173-180, 219. 

angular acceleration of connecting rod 
of, 176, 179. 

angular velocity of parts of, 172, 173, 
218. 

counterbalancing of, 248-256. 

dynamics of, 222-279. 

eccentric, 181. 

elementary combination of links of, 
10, 12, 163. 

fly-wheel, analysis of, 239-248. 

forces due to steam pressure in cylin- 
der of, 222-229. 

general description of chain, 163. 



3io 



INDEX 



Steam-engine — Cont. 

Gooch link, 207. 

governors, 256-279. 

inertia effects of reciprocating parts of, 
229-256. 

kinematics of, 163-221. 

link motions of, 198-207. 

multiple cylinder, 226. 

piston-crank chain of, 163-180. 

relative acceleration of parts of, 173- 
180, 219. 
position of parts of, 164-167, 219-221. 
velocity of parts of, 167-173. 

reversing gears of, 198-208. 

simple, 223-226. 

Stephenson link, 198-208. 

valve gearing of, 180-217. 
Stephenson link, 198-208. 

infinite blades in, 199-202. 

radius of link in, 203-205. 

short blades in, 202-207. 

valve diagrams for, 202-206. 
Stepped cones, 43-54. 

equations for crossed belt, 43. 
for open belt, 44-45. 

geometrical progression of speeds in, 

51-54- 
Palmer's graphical solution for, 47-49, 

53- 

Reuleaux's graphical solution £^,45-47. 

Sweet's graphical solution for, 49-50, 52. 
Straight line motions, 142-154. 

classification of, 142. 

conchoidal, 147-148. 

cycloidal, 142-147. 

Hart's, 153-154. 
- inversors, 150-154. 

lemniscate, 149. 

Peaucellier's, 151-153. 

Robert's, 147. 

Watt's, 150. 
Strength of spur gear teeth, 104. 
Sweet's graphical solution of geometrical 
progression, 51-54. 

graphical solution of the stepped cone, 

49-5°- 

T 

Tangential accelerative force moment, 

261, 276-278. 

components of forces at crank pin, 

228, 235. 



Tension, full theoretic, 264. 
Theoretic tension, full, 264. 
Throw of eccentric, 183. 
Toothed gearing, 66-141. 

angles of action, approach, and recess 
of, 72, 82, 92, 98. 

angular velocity, ratio in, 67-68. 

annular wheels, 76, 96. 

base circles of involute, 88. 

bevel, 114. 

circular spur, 72-105. 

curves of action of, 70. 

cycloidal system of, 73-85, 114. 

deduction of profile from curve of 
action of, 294-303. 

describing circle of cycloidal, 75-78. 

elliptic, 106-114. 

formulae for proportioning teeth of, 79, 
80. 

general considerations concerning, ^66. 

Grant's odontographic tables, 82, 90. 

hyperboloidal, 137-141. 

interchangeable wheels in, 77-78. 

interference in, 88-89, 97- 

involute system of, 85-93, TI 4- 

methods of draughting, 80-82, 90-91. 

non-circular spur, 105-113. 

obliquity of involute (standard) , 89. 

pin-tooth system of, 94-100. . 

pitch curves of, 67. 

surfaces of, 66, 114, 137. 

profiles of, 67, 294-303. 

racks, 74, jj, 86. 

skew, 117-141. 

sliding between teeth of, 69. 

spiral, 117-135- 

spur, 66-113. 

standard describing circle of cycloidal, 
78. 
dimensions of teeth of, 71. 
obliquity of involute, 89. 

strength of spur, 104-105. 

Tredgold's approximation, 116, 141. 

twisted, 100-104. 

worm, 135-137. 
Translation, rectilinear, 4, 141-154. 
Transmission of motion — 

directional relation not constant, 155- 

159- 
of rectilinear translation, 141-154. 
of rotation, 21-141. 



INDEX 



311 



Turning effort, 228, 240-242. 

mean, 229-239. 
Turning pairs, 8, 142. 
Twisted gears, 100-104. 

component pressures in, 104. 

U 

Uniplanar motion, 4. 
Universal joint, 25-31. 
angular acceleration of, 28-30. 
position of, 26. 
velocity of, 27-28. 
double, 31. 
Unstable governors, 266, 267. 



Valve, acceleration of, 219. 

distribution, 208, 211, 216. 

expansion, 208, 211, 216. 

gridiron, 208-211. 

inside lap of, 182. 

laps of, 182. 

lead of, 183. 

Meyer, 211-215. 

outside lap of, 182. 

plain slide, 180-208. 

position of, with respect to crank, 180- 
217. 
of, with respect to piston, 219-221. 

Thompson, 216-217. 

variable cut-off in plain slide, 195-208. 
in Meyer, 215. 
in Thompson, 217. 

velocity of, 218. 
Valve circles, critical points of, 188. 

gridiron, 209-210. 

Meyer, 213, 214, 216. 

plain slide, 184-189, 195-197, 202, 206. 
Valve diagrams. (See Valve Circles.) 
Valve gearing of steam-engine, 180-221. 

duty of, 180. 

problems in connection with, 190-195. 



Variable cut-off with constant admission 
in plain slide, 197. 
with constant lead in plain slide, 195- 

197. 
cut-off in Meyer valve, 215. 
in Stephenson link, 202. 
Thompson valve, 217. 
Velocity, angular, 17-18. 

angular, of connecting rod, 171-173. 

curve of fly-wheel, 242-243. 

curve of piston, 170. 

curve of valve, 218. 

determination of relative angular, 17. 

of relative linear, 12-17. 
of crank pin, 168. 
of points in piston-crank chain, 167- 

173- 
of valve, 218. 
of wrist pin, 169. 

W 

Watt's straight line motion, 150. 
Wire ropes, deflection at middle of span 
of, 57, 61. 

horizontal transmission by, 55-58. 

inclined transmission by, 58-62. 

sizes of sheaves for, 62. 

table of sizes of, etc., 56. 

tensions in, 54. 

transmission of rotation by, 55-62. 
Worm gears, 135-137. 

methods of hobbing, 136. 

tables for worm tools, 136-137. 
Wrapping connector, transmission of 

motion through, 36-37. 
Wrist pin, acceleration of, 173. 

forces active at, 227, 229-239. 

position of, relatively, to crank, 165-166. 

velocity of, 169-172. 



Zeuner valve diagram, 188, 210, 215. 
auxiliary circle in, 214. 



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A Laboratory flanual 

OF 

Physics and Applied Electricity. 

ARRANGED AND EDITED BY 

EDWARD L. NICHOLS, 

Professor of Physics in Cornell University. 
IN TWO VOLUMES. 

Vol. L JUNIOR COURSE IN GENERAL PHYSICS. 

BY 

ERNEST MERRITT and FREDERICK J. ROGERS. 

Cloth. $3.00. 

Vol. IL SENIOR COURSES AND OUTLINE OF 
ADVANCED WORK. 



GEORGE S. MOLER, FREDERICK BEDELL, HOMER J. HOTCHKISS, 
CHARLES P. MATTHEWS, and THE EDITOR. 

Cloth, pp. 444. $3.25. 



The first volume, intended for beginners, affords explicit directions adapted to a modern 
laboratory, together with demonstrations and elementary statements of principles. It is 
assumed that the student possesses some knowledge of analytical geometry and of the cal- 
culus. In the second volume more is left to the individual effort and to the maturer intel- 
ligence of the practicant. 

A large proportion of the students for whom primarily this Manual is intended, are pre- 
paring to become engineers, and especial attention has been devoted to the needs of that 
class of readers. In Vol. II., especially, a considerable amount of work in applied elec- 
tricity, in photometry, and in heat has been introduced. 

COMMENTS. 

"The work as a whole cannot be too highly commended. Its brief outlines of the 
various experiments are very satisfactory, its descriptions of apparatus are excellent; its 
numerous suggestions are calculated to develop the thinking and reasoning powers of the 
student. The diagrams are carefully prepared, and its frequent citations of original 
sources of information are of the greatest value." — Street Railway Journal. 

"The work is clearly and concisely written, the fact that it is edited by Professor Nichols 
being a sufficient guarantee of merit." — Electrical Engineering. 

"It will be a great aid to students. The notes of experiments and problems reveal 
much original work, and the book will be sure to commend itself to instructors." 

r- San Francisco Chro?iicie. 



THE MACMILLAN COMPANY 

66 FIFTH AVENUE, NEW YORK 
CHICAGO BOSTON SAN FRANCISCO ATLANTA 



THE ELEMENTS OF PHYSICS, 



BY 



EDWARD L. NICHOLS, B.S., Ph.D., 

Professor of Physics in Cornell University, 
AND 

WILLIAM S. FRANKLIN, M.S., 

Professor of Physics and Electrical Engineering at the Iowa Agricultural College, Ames, la, 
WITH NUMEROUS ILLUSTRATIONS. 

Part I. In Three Volumes: 

Vol. I. Mechanics and Heat . . Price $1.50 net. 

II. Electricity and Magnetism . " $1.90 net. 

III. Sound and Light ..." $1.50 net. 



It has been written with a view to providing a text-book which shall correspond with 
the increasing strength of the mathematical teaching in our university classes. In most of 
the existing text-books it appears to have been assumed that the student possesses so 
scanty a mathematical knowledge that he cannot understand the natural language of 
physics, i.e., the language of the calculus. Some authors, on the other hand, have assumed 
a degree of mathematical training such that their work is unreadable for nearly all under- 
graduates. 

The present writers having had occasion to teach large classes, the members of which 
were acquainted with the elementary principles of the calculus, have sorely felt the need of 
a text-book adapted to their students. The present work is an attempt on their part to 
supply this want. It is believed that in very many institutions a similar condition of affairs 
exists, and that there is a demand for a work of a grade intermediate between that of the 
existing elementary texts and the advanced manuals of physics. 

No attempt has been made in this work to produce a complete manual or compendium 
of experimental physics. The book is planned to be used in connection with illustrated 
lectures, in the course of which the phenomena are demonstrated and described. The 
authors have accordingly confined themselves to a statement of principles, leaving the 
lecturer to bring to notice the phenomena based upon them. In stating these principles, 
free use has been made of the calculus, but no demand has been made upon the student 
beyond that supplied by the ordinary elementary college courses on this subject. 

Certain parts of physics contain real and unavoidable difficulties. These have not been 
slurred over, nor have those portions of the subject which contain them been omitted. It 
has been thought more serviceable to the student and to the teacher who may have occa- 
sion to use the book to face such difficulties frankly, reducing the statements involving 
them to the simplest form which is compatible with accuracy. 

In a word, the Ele?nents of Physics is a book which has been written for use in such 
institutions as give their undergraduates a reasonably good mathematical training. It is 
intended for teachers who desire to treat their subject as an exact science, and who are 
prepared to supplement the brief subject-matter of the text by demonstration, illustration, 
and discussion drawn from the fund of their own knowledge. 



THE MACMILLAN COMPANY 

66 FIFTH AVENUE, NEW YORK 
CHICAGO BOSTON SAN FRANCISCO ATLANTA 



DEC 6 1902 



